NCERT Exemplar - Differential Equations Notes | EduRev

Mathematics (Maths) Class 12

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SHORT ANSWER TYPE QUESTIONS

Q.1. Find the solution ofNCERT Exemplar - Differential Equations Notes | EduRev
Ans.
The given differential equation is
NCERT Exemplar - Differential Equations Notes | EduRev
Separating the variables, we get
NCERT Exemplar - Differential Equations Notes | EduRev
Integrating both sides, we get
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev

⇒ – 2–y + 2–x = c log 2 

⇒ 2–x – 2–y = k [where c log 2 = k]

Q.2. Find the differential equation of all non vertical lines in a plane.
Ans.
Equation of all non vertical lines are y = mx + c
Differentiating with respect to x, we getNCERT Exemplar - Differential Equations Notes | EduRev
Again differentiating w.r.t. x we haveNCERT Exemplar - Differential Equations Notes | EduRev
Hence, the required equation isNCERT Exemplar - Differential Equations Notes | EduRev

Q.3. Given thatNCERT Exemplar - Differential Equations Notes | EduRevand y = 0 when x = 5. Find the value of x when y = 3.
Ans.
Given equation is
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Integrating both sides, we get
NCERT Exemplar - Differential Equations Notes | EduRev
Put y = 0 and x = 5
NCERT Exemplar - Differential Equations Notes | EduRev
∴ The equation becomesNCERT Exemplar - Differential Equations Notes | EduRev
Now putting y = 3, we get
NCERT Exemplar - Differential Equations Notes | EduRev
Hence the required value of x =NCERT Exemplar - Differential Equations Notes | EduRev

Q.4. Solve the differential equation (x2 – 1)NCERT Exemplar - Differential Equations Notes | EduRev

Ans.
Given differential equation is
NCERT Exemplar - Differential Equations Notes | EduRev
Dividing by (x2 – 1), we get
NCERT Exemplar - Differential Equations Notes | EduRev
It is a linear differential equation of first order and first degree.
NCERT Exemplar - Differential Equations Notes | EduRev
Integrating factor I.F. =NCERT Exemplar - Differential Equations Notes | EduRev
∴ Solution of the equation is
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Hence the required solution is NCERT Exemplar - Differential Equations Notes | EduRevNCERT Exemplar - Differential Equations Notes | EduRev

Q.5. Solve the differential equationNCERT Exemplar - Differential Equations Notes | EduRev
Ans.
Given equation isNCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Integrating both sides, we have
NCERT Exemplar - Differential Equations Notes | EduRev
⇒ log y =  x – x2 + log c
⇒ log y – log c = x – x2
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Hence, the required solution isNCERT Exemplar - Differential Equations Notes | EduRev

Q.6. Find the general solution ofNCERT Exemplar - Differential Equations Notes | EduRev
Ans.
Given equation isNCERT Exemplar - Differential Equations Notes | EduRev 
Here, P  = a and Q = emx
NCERT Exemplar - Differential Equations Notes | EduRev
Solution of equation is y x I.F =NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Hence the required solution isNCERT Exemplar - Differential Equations Notes | EduRev

Q.7. Solve the differential equationNCERT Exemplar - Differential Equations Notes | EduRev
Ans.
Given that:NCERT Exemplar - Differential Equations Notes | EduRev 
Put x  + y = t
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Integrating both sides, we have
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Hence, the required solution is ( x + c) . e xy + 1= 0.

Q.8. Solve: ydx – xdy = x2ydx.
Ans.
Given equation is ydx – xdy = x2y dx.
⇒ y dx - x2y dx = xdy
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Integrating both sides we get
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRevNCERT Exemplar - Differential Equations Notes | EduRev
Hence, the required solution is y =NCERT Exemplar - Differential Equations Notes | EduRev

Q.9. Solve the differential equationNCERT Exemplar - Differential Equations Notes | EduRev= 1 + x + y2 + xy2, when y = 0, x = 0.
Ans.
Given equation is
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Integrating both sides, we get
NCERT Exemplar - Differential Equations Notes | EduRev
Put x  = 0 and y = 0, we get tan–1(0) = 0 + 0 + c
⇒ c = 0
NCERT Exemplar - Differential Equations Notes | EduRev
Hence, the required solution isNCERT Exemplar - Differential Equations Notes | EduRev

Q.10. Find the general solution ofNCERT Exemplar - Differential Equations Notes | EduRev
Ans.
Given equation isNCERT Exemplar - Differential Equations Notes | EduRev 
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
HereNCERT Exemplar - Differential Equations Notes | EduRev
∴ Integrating factor I.F. =NCERT Exemplar - Differential Equations Notes | EduRev
So the solution of the equation is
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
So x = y3 + cy = y(y2 + c)
Hence, the required solution is x = y (y2 + c).

Q.11. If y(x) is a solution ofNCERT Exemplar - Differential Equations Notes | EduRev= – cos x and y (0) = 1, then find the value ofNCERT Exemplar - Differential Equations Notes | EduRev
Ans.
Given equation is
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Integrating both sides, we get
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Put x  = 0 and y = 1, we get
(1 + 1) (2 + sin 0) = c ⇒ 4 = c
∴ equation is (1 + y) (2 + sin x) = 4
Now putNCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Hence, the required solution isNCERT Exemplar - Differential Equations Notes | EduRev

Q.12. If y(t) is a solution of (1 + t)NCERT Exemplar - Differential Equations Notes | EduRevand y (0) = – 1, then show that

y (1) =NCERT Exemplar - Differential Equations Notes | EduRev
Ans.
Given equation is
NCERT Exemplar - Differential Equations Notes | EduRev
Here,NCERT Exemplar - Differential Equations Notes | EduRev
∴ Integrating factor I.F.NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
∴ I.F. = e–t . (1 + t)
Required solution of the given differential equation is
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Put t = 0 and y = –1 [∵ y(0) = –1]
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
So the equation becomes
ye–t (1 + t) = –e–t
Now put t = 1
∴ y . e–1 (1 + 1) = –e–1
NCERT Exemplar - Differential Equations Notes | EduRev
Hence y (1) =NCERT Exemplar - Differential Equations Notes | EduRevis verified.

Q.13. Form the differential equation having y = (sin–1x)2 + Acos–1x + B, where A and B are arbitrary constants, as its general solution.
Ans.
Given equation is y = (sin–1x)2 + A cos–1x + B
NCERT Exemplar - Differential Equations Notes | EduRev
Multiplying both sides byNCERT Exemplar - Differential Equations Notes | EduRevwe get
NCERT Exemplar - Differential Equations Notes | EduRev
Again differentiating w.r.t x, we get
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Multiplying both sides byNCERT Exemplar - Differential Equations Notes | EduRev, we get
NCERT Exemplar - Differential Equations Notes | EduRev
Which is the required differential equation.

Q.14. Form the differential equation of all circles which pass through origin and whose centres lie on y-axis.
Ans.
Equation of circle which passes through the origin and whose centre lies on y-axis is
(x – 0)2 + (y – a)2 = a2
⇒ x2  + y2 + a2 – 2ay = a2
⇒ x2  + y2 – 2ay = 0 ...(i)
Differentiating both sides w.r.t. x we get
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Putting the value of a in eq. (i), we get
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Hence, the required differential equation is
NCERT Exemplar - Differential Equations Notes | EduRev

Q.15. Find the equation of a curve passing through origin and satisfying the differential equationNCERT Exemplar - Differential Equations Notes | EduRev
Ans.
Given equation is
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Here,NCERT Exemplar - Differential Equations Notes | EduRev
Integrating factor I.F. =NCERT Exemplar - Differential Equations Notes | EduRev
∴ Solution is
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev...(i)
Since the curve is passing through origin i.e., (0, 0)
∴ Put y = 0 and x = 0 in eq. (i)
NCERT Exemplar - Differential Equations Notes | EduRev
∴ Equation isNCERT Exemplar - Differential Equations Notes | EduRev
Hence, the required solution isNCERT Exemplar - Differential Equations Notes | EduRev

Q.16. Solve :NCERT Exemplar - Differential Equations Notes | EduRev
Ans.
Given equation isNCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Put y  = vx  [∵ it is a homogeneous differential equation]
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Integrating both sides, we get
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Hence, the required solution isNCERT Exemplar - Differential Equations Notes | EduRev

Q.17. Find the general solution of the differential equation (1 + y2) + 
(x – etan–1y)NCERT Exemplar - Differential Equations Notes | EduRev
Ans.
Given equation is
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Here,NCERT Exemplar - Differential Equations Notes | EduRev
∴ Integrating factor I.F. =NCERT Exemplar - Differential Equations Notes | EduRev
∴ Solution is
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Hence, this is the required general solution.

Q.18. Find the general solution of y2dx + (x2 – xy + y2) dy = 0.
Ans.
The given equation is y2 dx + ( x2 - xy+ y2 ) dy =0 .
⇒ y2dx = – (x2 – xy + y2) dy
NCERT Exemplar - Differential Equations Notes | EduRev
Since it is a homogeneous differential equation
NCERT Exemplar - Differential Equations Notes | EduRev
So,NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Integrating both sides, we get
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Hence the required solution isNCERT Exemplar - Differential Equations Notes | EduRev

Q.19. Solve: (x + y) (dx – dy) = dx + dy. [Hint: Substitute x + y = z after seperating dx and dy]
Ans.
Given differential equation is (x + y) (dx – dy) = dx + dy
⇒ (x + y) dx – (x + y) dy = dx + dy
⇒ – (x + y) dy – dy = dx – (x + y) dx
⇒ – (x + y + 1) dy = – (x + y – 1) dx
NCERT Exemplar - Differential Equations Notes | EduRev
Put x + y = z
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
So,NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Integrating both sides, we get
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
∴ x + y = c . ex – y
Hence, the required solution is x + y = c . ex – y.

Q.20. Solve: 2(y + 3) – xyNCERT Exemplar - Differential Equations Notes | EduRevgiven that y (1) = – 2.
Ans.
Given differential equation is
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Integrating both sides, we get
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Put x = 1, y = –2
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev

⇒ – 1 – 0 = 0 + c [∵ log (1) = 0]
∴ c = –1
∴ equation is
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Hence, the required solution is x2 (y + 3)3 = ey + 2

Q.21. Solve the differential equation dy = cos x (2 – y cosec x) dx given that y = 2 whenNCERT Exemplar - Differential Equations Notes | EduRev
Ans.
The given differential equation is
dy = cos x (2 – y cosec x) dx
NCERT Exemplar - Differential Equations Notes | EduRev
Here, P = cot x and Q = 2 cos x.
∴ Integrating factor I.F. =NCERT Exemplar - Differential Equations Notes | EduRev
∴ Required solution is y × I.F =NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Put x =NCERT Exemplar - Differential Equations Notes | EduRevand y = 2, we get
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
∴ The equation is y sin x =NCERT Exemplar - Differential Equations Notes | EduRev

Q.22. Form the differential equation by eliminating A and B in Ax2 + By2 = 1.
Ans.
Given that Ax2 + By2 = 1
Differentiating w.r.t. x, we get
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Differentiating both sides again w.r.t. x, we have
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Hence, the required equation is
xy.y ″ + x.( y ′)2- y.y′ = 0

Q.23. Solve the differential equation (1 + y2) tan–1x dx + 2y (1 + x2) dy = 0.
Ans.
Given differential equation is
(1 + y2) tan–1x dx + 2y (1 + x2) dy = 0
⇒ 2y (1 + x2) dy = –(1 + y2) . tan–1x . dx
NCERT Exemplar - Differential Equations Notes | EduRev
Integrating both sides, we get
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Which is the required solution.

Q.24. Find the differential equation of system of concentric circles with centre (1, 2).
Ans.
 Family of concentric circles with centre (1, 2) and radius ‘r’ is
(x – 1)2 + (y – 2)2 = r2 
Differentiating both sides w.r.t., x we get
NCERT Exemplar - Differential Equations Notes | EduRev
Which is the required equation.


LONG ANSWER TYPE QUESTIONS

Q.25. Solve :NCERT Exemplar - Differential Equations Notes | EduRev
Ans.
The given differential equation is
NCERT Exemplar - Differential Equations Notes | EduRev 
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Here, P =NCERT Exemplar - Differential Equations Notes | EduRevand Q = (sin x + log x)
Integrating factor I.F. =NCERT Exemplar - Differential Equations Notes | EduRev
∴ Solution is
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev...(1)
Let I
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Now from eq (1) we get,
NCERT Exemplar - Differential Equations Notes | EduRev
Hence, the required solution is
NCERT Exemplar - Differential Equations Notes | EduRev

Q.26. Find the general solution of (1 + tany) (dx – dy) + 2xdy = 0.
Ans.
Given that: (1 + tan y) (dx – dy) + 2xdy = 0
⇒ (1 + tan y) dx – (1 + tan y) dy + 2xdy = 0
⇒ (1 + tan y) dx – (1 + tan y – 2x) dy = 0
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Here, P =NCERT Exemplar - Differential Equations Notes | EduRev
Integrating factor I.F.
NCERT Exemplar - Differential Equations Notes | EduRev
So, the solution is x × I.F. =NCERT Exemplar - Differential Equations Notes | EduRev
⇒ x . ey (sin y + cos y) =NCERT Exemplar - Differential Equations Notes | EduRev
⇒ x . ey (sin y + cos y) = ey . sin y + c
NCERT Exemplar - Differential Equations Notes | EduRev
⇒ x(sin y + cos y) = sin y + c . e–y
Hence, the required solution is x(sin y + cos y) = sin y + c . e–y.

Q.27. Solve :NCERT Exemplar - Differential Equations Notes | EduRev [Hint: Substitute x + y = z]
Ans.
Given that :NCERT Exemplar - Differential Equations Notes | EduRev
Put x + y = v, on differentiating w.r.t. x, we get,
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev-1 = cos v + sin v
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Integrating both sides, we have
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
PutNCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Hence, the required solution is
NCERT Exemplar - Differential Equations Notes | EduRev= x + K [c – log 2 = K]

Q.28. Find the general solution of NCERT Exemplar - Differential Equations Notes | EduRev
Ans.
Given equation isNCERT Exemplar - Differential Equations Notes | EduRev 
Here, P = –3 and Q = sin 2 x
∴ Integrating factor I.F. =NCERT Exemplar - Differential Equations Notes | EduRev
∴ Solution is
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
LetNCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
∴ The equation becomes
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Hence, the required solution is
NCERT Exemplar - Differential Equations Notes | EduRev

Q.29. Find the equation of a curve passing through (2, 1) if the slope of the tangent to the curve at any point (x, y) isNCERT Exemplar - Differential Equations Notes | EduRev
Ans.
Given that the slope of tangent to a curve at (x, y) is
NCERT Exemplar - Differential Equations Notes | EduRev
It is a homogeneous differential equation
So, put y = vxNCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Integrating both sides, we get
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Since, the curve is passing through the point (2, 1)
NCERT Exemplar - Differential Equations Notes | EduRev
Hence, the required equation is
NCERT Exemplar - Differential Equations Notes | EduRev

Q.30. Find the equation of the curve through the point (1, 0) if the slope of the tangent to the curve at any point (x, y) isNCERT Exemplar - Differential Equations Notes | EduRev
Ans.
Given that the slope of the tangent to the curve at (x, y) is
NCERT Exemplar - Differential Equations Notes | EduRev
Integrating both sides, we have
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev[making perfect square]
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Since, the line is passing through the point (1, 0), then
(0 – 1) (1 + 1) = c(1) ⇒ c = 2.
Hence, the required solution is (y – 1) (x + 1) = 2x.

Q.31. Find the equation of a curve passing through origin if the slope of the tangent to the curve at any point (x, y) is equal to the square of the difference of the abcissa and ordinate of the point.
Ans.
Here, slope of the tangent of the curve =NCERT Exemplar - Differential Equations Notes | EduRev
and the difference between the abscissa and ordinate = x – y.
∴ As per the condition,NCERT Exemplar - Differential Equations Notes | EduRev
Put x – y = v
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
∴ the equation becomes
NCERT Exemplar - Differential Equations Notes | EduRev
Integrating both sides, we get
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev...(1)
Since, the curve is passing through (0, 0)
thenNCERT Exemplar - Differential Equations Notes | EduRev
∴ On putting c = 0 in eq. (1) we get
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
⇒ (1 + x – y) = e2x (1 – x + y)
Hence, the required equation is (1 + x – y) = e2x (1 – x + y).

Q.32. Find the equation of a curve passing through the point (1, 1). If the tangent drawn at any point P (x, y) on the curve meets the co-ordinate axes at A and B such that P is the mid-point of AB.
Ans.

Let P (x, y) be any point on the curve and AB be the tangent to the given curve at P.

P is the mid point of AB (given)
∴ Coordinates of A and B are (2x, 0) and (0, 2y) respectively.
∴ Slope of the tangent
NCERT Exemplar - Differential Equations Notes | EduRev
AB =NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Integrating both sides, we get
NCERT Exemplar - Differential Equations Notes | EduRev

⇒ log y + log x = log c ⇒ log yx = log c
∴ yx = c
Since, the curve passes through (1, 1)
∴ 1 × 1 = c ∴ c = 1
⇒ yx = 1
Hence, the required equation is xy = 1.

Q.33. Solve :NCERT Exemplar - Differential Equations Notes | EduRev 
Ans.
Given that:NCERT Exemplar - Differential Equations Notes | EduRev 
NCERT Exemplar - Differential Equations Notes | EduRev
Since, it is a homogeneous differential equation.
∴ PutNCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Integrating both sides, we get
NCERT Exemplar - Differential Equations Notes | EduRev
Put log v = t on L.H.S.
NCERT Exemplar - Differential Equations Notes | EduRevdv = dt
NCERT Exemplar - Differential Equations Notes | EduRev
log|t| = log|x| + log c
⇒ log |log v| = log xc ⇒ log v = xc
NCERT Exemplar - Differential Equations Notes | EduRev
Hence, the required solution is logNCERT Exemplar - Differential Equations Notes | EduRev


OBJECTIVE TYPE QUESTIONS

Q.34. The degree of the differential equationNCERT Exemplar - Differential Equations Notes | EduRev
(a) 1 
(b) 2 
(c) 3 
(d) not defined
Ans. (d)
Solution.
The degree of the given differential equation is not defined
because the value of sinNCERT Exemplar - Differential Equations Notes | EduRevon expansion will be in increasing power of
NCERT Exemplar - Differential Equations Notes | EduRev
Hence, the correct option is (d).

Q.35. The degree of the differential equationNCERT Exemplar - Differential Equations Notes | EduRev
(a) 4
(b) 3/2

(c) not defined
(d) 2

Ans. (d)
Solution.
The given differential equation is
NCERT Exemplar - Differential Equations Notes | EduRev
Squaring both sides, we have
NCERT Exemplar - Differential Equations Notes | EduRev
So, the degree of the given differential equation is 2.
Hence, the correct option is (d).

Q.36. T he order and degree of the differential equationNCERT Exemplar - Differential Equations Notes | EduRev respectively are
(a) 2 and not defined 
(b) 2 and 2 
(c) 2 and 3 
(d) 3 and 3
Ans. (a)
Solution.

Given differential equation is
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRevNCERT Exemplar - Differential Equations Notes | EduRev
Since the degree ofNCERT Exemplar - Differential Equations Notes | EduRevis in fraction.
So, the degree of the differential equation is not defined as the order is 2.
Hence, the correct option is (a).

Q.37. If y = e–x (Acosx + Bsinx), then y is a solution of
(a)NCERT Exemplar - Differential Equations Notes | EduRev
(b)NCERT Exemplar - Differential Equations Notes | EduRev
(c)NCERT Exemplar - Differential Equations Notes | EduRev
(d)NCERT Exemplar - Differential Equations Notes | EduRev
Ans. (c)
Solution.

Given equation is y = e–x (A cos x + B sin x)
Differentiating both sides, w.r.t. x, we get
NCERT Exemplar - Differential Equations Notes | EduRev
Again differentiating w.r.t. x, we get
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Hence, the correct option is (c)

Q.38. The differential equation for y = Acos αx + Bsin αx, where A and B are arbitrary constants is
(a)NCERT Exemplar - Differential Equations Notes | EduRev
(b)NCERT Exemplar - Differential Equations Notes | EduRev
(c)NCERT Exemplar - Differential Equations Notes | EduRev
(d)NCERT Exemplar - Differential Equations Notes | EduRev
Ans. (b)
Solution.

Given equation is : y = A cos α x + B sin α x
Differentiating both sides w.r.t. x, we have
NCERT Exemplar - Differential Equations Notes | EduRev= –A sin α x . α + B cos α x . α
= –A α sin α x + B α cos α x
Again differentiating w.r.t. x, we get
NCERT Exemplar - Differential Equations Notes | EduRev= –A α2 cos α x – B α2 sin α x
NCERT Exemplar - Differential Equations Notes | EduRev= – α2 (A cos α x + B sin α x)
NCERT Exemplar - Differential Equations Notes | EduRev= – α2 y ⇒NCERT Exemplar - Differential Equations Notes | EduRev
Hence, the correct option is (b).

Q.39. Solution of differential equation xdy – ydx = 0 represents :
(a) a rectangular hyperbola 
(b) parabola whose vertex is at origin 
(c) straight line passing through origin 
(d) a circle whose centre is at origin.
Ans. (c)
Solution.

The given differential equation is
xdy – ydx = 0
NCERT Exemplar - Differential Equations Notes | EduRev
Integrating both sides, we get
NCERT Exemplar - Differential Equations Notes | EduRev

 log y = log x + log c ⇒ log y = log xc
⇒ y = xc  which is  a straight line passing through the origin.      
Hence, the correct answer is (c).

Q.40. Integrating factor of the differential equation cosxNCERT Exemplar - Differential Equations Notes | EduRev
(a) cos x 
(b) tan 
(c) sec 
(d) sin x
Ans. (c)
Solution.

The given differential equation is
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Here, P = tan x and Q = sec x
 Integrating factor =NCERT Exemplar - Differential Equations Notes | EduRev
 Hence, the correct option is (c).

Q.41. Solution of the differential equation tany sec2x dx + tanx sec2ydy = 0 is :
(a) tan x + tan y = k 
(b) tan x  tan y = k 
(c)NCERT Exemplar - Differential Equations Notes | EduRev
(d) tanx . tany = k
Ans. (d)
Solution.

The given differential equation is      
tan y sec2 x dx + tan x sec2 y dy = 0
 tan x sec2 y dy = - tan y sec2 x dx
NCERT Exemplar - Differential Equations Notes | EduRev
Integrating both sides, we get
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev 

Q.42. Family y = Ax + A3 of curves is represented by the differential 
equation of degree
(a) 1 
(b) 2 
(c) 3 
(d) 4
Ans. (a)
Solution.

Given equation is y = Ax + A3 
Differentiating both sides, we get
NCERT Exemplar - Differential Equations Notes | EduRev= A which has degree 1.

Hence, the correct answer is (a).

Q.43. Integrating factor ofNCERT Exemplar - Differential Equations Notes | EduRev

(a) x
(b) log x

(c) 1/x

(d) – x
Ans. (c)
Solution.

The given differential equation is
NCERT Exemplar - Differential Equations Notes | EduRev
Here,NCERT Exemplar - Differential Equations Notes | EduRev
So, integrating factor =NCERT Exemplar - Differential Equations Notes | EduRev
Hence, the correct option is (c).

Q.44. Solution ofNCERT Exemplar - Differential Equations Notes | EduRevy (0) = 1 is given by
(a) xy = – ex 
(b) xy = – e–x 
(c) xy = – 1 
(d) y = 2 ex – 1
Ans. (d)
Solution.

The given differential equation is
NCERT Exemplar - Differential Equations Notes | EduRev
Here, P = –1, Q = 1
∴ Integrating factor, I.F. =NCERT Exemplar - Differential Equations Notes | EduRev
So, the solution is
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev

⇒ y . e–x = – e–x + c
Put x = 0, y = 1
⇒ 1 . e0 = – e0 + c

⇒ 1 = –1 + c ∴ c = 2

So the equation is  y . e–x = –e–x + 2
⇒ y = – 1 + 2ex = 2ex – 1
Hence, the correct option is (d).

Q.45. The number of solutions ofNCERT Exemplar - Differential Equations Notes | EduRevwhen y (1) = 2 is

(a) none
(b) one
(c) two
(d) infinite

Ans. (b)
Solution.

The given differential equation isNCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Integrating both sides, we get
NCERT Exemplar - Differential Equations Notes | EduRev
⇒ log (y + 1) = log (x – 1) + log c
⇒ log (y + 1) – log (x – 1) = log c
NCERT Exemplar - Differential Equations Notes | EduRev
Put x = 1 and y = 2
NCERT Exemplar - Differential Equations Notes | EduRev∴ c = ∞
NCERT Exemplar - Differential Equations Notes | EduRev⇒ x – 1 = 0 ⇒ x = 1
Hence, the correct option is (b).

Q.46. Which of the following is a second order differential equation?
(a) (y′)2 + x = y2 
(b) y′y′′+ y = sin x 
(c) y″ + (y'')2 + y = 0 
(d) y′ = y2
Ans. (b)
Solution.

Second order differential equation is y′y″+ y = sin x
Hence, the correct option is (b).

Q.47. Integrating factor of the differential equation (1 – x2)NCERT Exemplar - Differential Equations Notes | EduRev
(a) –x
(b)NCERT Exemplar - Differential Equations Notes | EduRev
(c)NCERT Exemplar - Differential Equations Notes | EduRev
(d)NCERT Exemplar - Differential Equations Notes | EduRev
Ans. (c)
Solution.

The given differential equation is
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Here,NCERT Exemplar - Differential Equations Notes | EduRev
∴ Integrating factor
NCERT Exemplar - Differential Equations Notes | EduRev
Hence, the correct option is (c).

Q.48. tan–1 x + tan–1 y = c is the general solution of the differential equation:
(a)NCERT Exemplar - Differential Equations Notes | EduRev
(b)NCERT Exemplar - Differential Equations Notes | EduRev
(c) (1 + x2) dy + (1 + y2) dx = 0
(d) (1 + x2) dx + (1 + y2) dy = 0

Ans. (c)
Solution.

Given equation is tan–1x + tan–1y = c
Differentiating w.r.t. x, we have
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev

⇒(1 + x2) dy = – (1 + y2) dx
⇒ (1 + x2) dy + (1 + y2) dx = 0
Hence the correct option is (c).

Q.49. The differential equation yNCERT Exemplar - Differential Equations Notes | EduRevrepresents :
(a) Family of hyperbolas 
(b) Family of parabolas 
(c) Family of ellipses 
(d) Family of circles
Ans. (d)
Solution.

Given differential equation is
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev= c – x ⇒ y dy = (c – x) dx
∴ Integrating both sides, we get
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev

⇒ x2 + y2 – 2cx = 2k which is a family of circles.

Hence, the correct option is (d).


Q.50. The general solution of ex cosy dx – ex siny dy = 0 is :
(a) ex cos y = k 
(b) ex sin y = k 
(c) ex = k cos y 
(d) ex = k sin y
Ans. (a)
Solution.

The given differential equation is
ex cos y dx – ex sin y dy = 0
⇒ ex (cos y dx – sin y dy) = 0
 cos y dx - sin y dy = 0 [∵ ex ≠ 0]
 sin y dy = cos y dxNCERT Exemplar - Differential Equations Notes | EduRev
Integrating both sides, we have
NCERT Exemplar - Differential Equations Notes | EduRev
⇒ - log |cos y| = x + log k ⇒NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRevNCERT Exemplar - Differential Equations Notes | EduRev
Hence, the correct option is (a).

Q.51. The degree of the differential equationNCERT Exemplar - Differential Equations Notes | EduRev    is 
(a) 1 
(b) 2 
(c) 3 
(d) 5
Ans. (a)
Solution.

The degree of the given differential equation is 1 as the power of the highest order is 1.
Hence, the correct option is (a).

Q.52. The solution ofNCERT Exemplar - Differential Equations Notes | EduRevis 
(a) y = ex (x – 1) 
(b) y = xe–x 
(c) y = x e–x + 1 
(d) y = (x + 1) e–x
Ans. (b)
Solution.

The given differential equation is
NCERT Exemplar - Differential Equations Notes | EduRev
Since, it is a linear differential equation
NCERT Exemplar - Differential Equations Notes | EduRev
So, the solution is
NCERT Exemplar - Differential Equations Notes | EduRev
Put x = 0, y = 0, we have 0 = 0 + c ∴ c = 0
So, the solution is y ex = x ⇒ y = x . e–x
Hence, the correct option is (b).

Q.53. Integrating factor of the differential equation
NCERT Exemplar - Differential Equations Notes | EduRev
(a) cos x 
(b) sec x 
(c) ecos x 
(d) esec x
Ans. (b)
Solution.

Given differential equation is
NCERT Exemplar - Differential Equations Notes | EduRev
Here, P = tan x and Q = sec x
NCERT Exemplar - Differential Equations Notes | EduRev
Hence, the correct option is (b).

Q.54. The solution of the differential equationNCERT Exemplar - Differential Equations Notes | EduRev
(a) y = tan–1
(b) y – x = k (1 + xy) 
(c) x = tan–1
(d) tan (xy) = k
Ans. (b)
Solution.

The given differential equation is
NCERT Exemplar - Differential Equations Notes | EduRevNCERT Exemplar - Differential Equations Notes | EduRev
Integrating both sides, we get
NCERT Exemplar - Differential Equations Notes | EduRev
⇒tan–1 y = tan–1 x + c ⇒ tan-1 y - tan-1 x = c
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
⇒ y – x = k (1 + xy)
Hence, the correct option is (b).

Q.55. The integrating factor of the differential equationNCERT Exemplar - Differential Equations Notes | EduRev
(a)NCERT Exemplar - Differential Equations Notes | EduRev
(b)NCERT Exemplar - Differential Equations Notes | EduRev
(c) xe
(d) ex

Ans. (b)
Solution.

The given differential equation is
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Here, P =NCERT Exemplar - Differential Equations Notes | EduRev
∴ Integrating factor I.F.
NCERT Exemplar - Differential Equations Notes | EduRev
Hence, the correct option is (b).

Q.56. y = aemx + be–mx satisfies which of the following differential equation?
(a)NCERT Exemplar - Differential Equations Notes | EduRev
(b)NCERT Exemplar - Differential Equations Notes | EduRev
(c)NCERT Exemplar - Differential Equations Notes | EduRev
(d)NCERT Exemplar - Differential Equations Notes | EduRev
Ans. (c)
Solution.

The given equation is  y = aemx + be–mx 
On differentiation, we getNCERT Exemplar - Differential Equations Notes | EduRev
Again differentiating w.r.t., we have
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Hence, the correct option is (c).

Q.57. The solution of the differential equation cosx siny dx + sinx cosy dy = 0 
is :
(a)NCERT Exemplar - Differential Equations Notes | EduRev
(b) sin x  sin y = c 
(c) sin x + sin y = c 
(d) cos x cos y = c
Ans. (b)
Solution.

The given differential equation is
cos x sin y dx + sin x cos y dy = 0
⇒ sin x cos y d y = – cos x sin y dx
NCERT Exemplar - Differential Equations Notes | EduRev
Integrating both sides, we have
NCERT Exemplar - Differential Equations Notes | EduRev
⇒ log |sin y| = – log |sin x| + log c
⇒ log |sin y| + log |sin x| = log c
⇒ log |sin y . sin x| = log c ⇒ sin x sin y = c
Hence, the correct option is (b).

Q.58. The solution ofNCERT Exemplar - Differential Equations Notes | EduRev
(a)NCERT Exemplar - Differential Equations Notes | EduRev
(b) y = xex + cx
(c) y = x . ex + k
(d)NCERT Exemplar - Differential Equations Notes | EduRev

Ans. (a)
Solution.

The given differential equation isNCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Here P NCERT Exemplar - Differential Equations Notes | EduRev
∴ Integrating factor I.F. =NCERT Exemplar - Differential Equations Notes | EduRev
So, the solution is
NCERT Exemplar - Differential Equations Notes | EduRev
Hence, the correct option is (a).

Q.59. The differential equation of the family of curves x2 + y2 – 2ay = 0, where a is arbitrary constant, is:
(a)NCERT Exemplar - Differential Equations Notes | EduRev
(b)NCERT Exemplar - Differential Equations Notes | EduRev
(c)NCERT Exemplar - Differential Equations Notes | EduRev
(d)NCERT Exemplar - Differential Equations Notes | EduRev
Ans. (a)
Solution.

The given equation is
x2 + y2 – 2ay = 0 ..(1)
Differentiating w.r.t. x, we have
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRevNCERT Exemplar - Differential Equations Notes | EduRev
Putting the value of a in eq. (1) we get
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
∴ Hence the correct option is (a).

Q.60. Family y = Ax + A3 of curves will correspond to a differential equation of order
(a) 3 
(b) 2 
(c) 1 
(d) not defined
Ans. (b)
Solution.

The given equation is
y = Ax + A3 
Differentiating both sides, we getNCERT Exemplar - Differential Equations Notes | EduRev
Again differentiating both sides, we haveNCERT Exemplar - Differential Equations Notes | EduRev

So the order of the differential equation is 2.

Hence, the correct option is (b).

Q.61. The general solution ofNCERT Exemplar - Differential Equations Notes | EduRev

(a)NCERT Exemplar - Differential Equations Notes | EduRev
(b)NCERT Exemplar - Differential Equations Notes | EduRev
(c)NCERT Exemplar - Differential Equations Notes | EduRev
(d)NCERT Exemplar - Differential Equations Notes | EduRev
Ans. (c)
Solution.

The given differential equation is
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Integrating both sides, we have
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Hence, the correct option is (c).

Q.62. The curve for which the slope of the tangent at any point is equal to the ratio of the abcissa to the ordinate of the point is
(a) an ellipse 
(b) parabola 
(c) circle 
(d) rectangular hyperbola
Ans. (d)
Solution.

Since, the slope of the tangent to the curve = x : y
NCERT Exemplar - Differential Equations Notes | EduRev
Integrating both sides, we getNCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
⇒ y2 – x2 = 2c = k which is rectangular hyperbola.

Hence, the correct option is (d).

Q.63. The general solution of the differential equation
NCERT Exemplar - Differential Equations Notes | EduRev
(a)NCERT Exemplar - Differential Equations Notes | EduRev
(b)NCERT Exemplar - Differential Equations Notes | EduRev
(c)NCERT Exemplar - Differential Equations Notes | EduRev
(d)NCERT Exemplar - Differential Equations Notes | EduRev
Ans. (c)
Solution.

The given differential equation is
NCERT Exemplar - Differential Equations Notes | EduRev
Since it is linear differential equation where P = –x and Q =NCERT Exemplar - Differential Equations Notes | EduRev
∴ Integrating factor I.F. =NCERT Exemplar - Differential Equations Notes | EduRev
So, the solution is
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Hence the correct option is (c).

Q.64. The solution of the equation (2y – 1) dx – (2x + 3)dy = 0 is :
(a)NCERT Exemplar - Differential Equations Notes | EduRev
(b)NCERT Exemplar - Differential Equations Notes | EduRev
(c)NCERT Exemplar - Differential Equations Notes | EduRev
(d)NCERT Exemplar - Differential Equations Notes | EduRev
Ans. (c)
Solution.

The given differential equation is
(2y – 1) dx – (2x + 3) dy = 0 ⇒ (2x + 3) dy = (2y - 1) dx
NCERT Exemplar - Differential Equations Notes | EduRev
Integrating both sides, we get
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
⇒log |2y – 1| = log |2x + 3| + 2 log c
⇒ log |2y – 1| – log |2x + 3| = log c2
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Hence, the correct option is (c).

Q.65. The differential equation for which y = acosx + bsinx is a solution, is :
(a)NCERT Exemplar - Differential Equations Notes | EduRev
(b)NCERT Exemplar - Differential Equations Notes | EduRev
(c)NCERT Exemplar - Differential Equations Notes | EduRev
(d)NCERT Exemplar - Differential Equations Notes | EduRev
Ans. (a)
Solution.

The given equation is
y = a cos x + b sin x
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Hence, the correct option is (a).


Q.66. The solution ofNCERT Exemplar - Differential Equations Notes | EduRev
(a) y = e–x (x – 1) 
(b) y = x . ex 
(c) y = x e–x + 1 
(d) y = x . e–x
Ans. (d)
Solution.

The given differential equation isNCERT Exemplar - Differential Equations Notes | EduRev
Since, it is a linear differential equation then P = 1 and Q = e–x
Integrating factor I.F. =NCERT Exemplar - Differential Equations Notes | EduRev
∴ Solution is
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev

Put y = 0 and x = 0
∴ 0 = 0 + c ∴ c = 0
∴ equation is y × ex = x
So y = x . e–x 
Hence, the correct option is (d).

Q.67. The order and degree of the differential equation
NCERT Exemplar - Differential Equations Notes | EduRev
(a) 1, 4 
(b) 3, 4 
(c) 2, 4 
(d) 3, 2
Ans. (d)
Solution.

The given differential equation is
NCERT Exemplar - Differential Equations Notes | EduRev
Here the highest derivative isNCERT Exemplar - Differential Equations Notes | EduRev
∴ the order of the differential equation is 3
and since, the power of highest order is 2
∴ its degree is 2
Hence, the correct option is (d).

Q.68. The order and degree of the differential equation
NCERT Exemplar - Differential Equations Notes | EduRev
(a)NCERT Exemplar - Differential Equations Notes | EduRev
(b) 2, 3 
(c) 2, 1 
(d) 3, 4
Ans. (c)
Solution.

The given differential equation is
NCERT Exemplar - Differential Equations Notes | EduRev
Here, the highest derivative is 2,
∴ order = 2
and the power of the highest derivative is 1
∴ degree = 1
Hence, the correct option is (c).

Q.69. The differential equation of the family of curves y2 = 4a (x + a) is :
(a)NCERT Exemplar - Differential Equations Notes | EduRev
(b)NCERT Exemplar - Differential Equations Notes | EduRev
(c)NCERT Exemplar - Differential Equations Notes | EduRev
(d)NCERT Exemplar - Differential Equations Notes | EduRev
Ans. (d)
Solution.

The given equation of family of curves is
y2 = 4a (x + a)

⇒ y2 = 4ax + 4a2 ...(1)
Differentiating both sides, w.r.t. x, we get
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Now, putting the value of a in eq. (1) we get
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Hence, the correct option is (d).

Q.70. Which of the following is the general solution ofNCERT Exemplar - Differential Equations Notes | EduRev
(a) y = (Ax + B) . ex 
(b) y = (Ax + B) e–x 
(c) y = Aex + Be–x 
(d) y = A cos x + B sin x
Ans. (a)
Solution.

The given differential equation is
NCERT Exemplar - Differential Equations Notes | EduRev
Since the above equation is of second order and first degree
∴ D2y – 2Dy + y = 0, where D =NCERT Exemplar - Differential Equations Notes | EduRev
⇒ (D2 - 2D + 1) y = 0
∴ auxiliary equation is
m2 - 2m + 1 = 0 ⇒ (m - 1)2 = 0 ⇒ m = 1, 1
If the roots of Auxiliary equation are real and equal say (m)
then CF = (c1 x + c2) . emx
∴ CF = (Ax + B) ex
So y = (Ax + B) . ex
Hence, the correct option is (a).

Q.71. General solution ofNCERT Exemplar - Differential Equations Notes | EduRev
(a) y sec x = tan x + c 
(b) y tan x = sec x + c 
(c) tan x = y tan x + c 
(d) x sec x = tan y + c
Ans. (a)
Solution.

The given differential equation isNCERT Exemplar - Differential Equations Notes | EduRev
Since, it is a linear differential equation
∴ P = tan x and Q = sec x
Integrating factor I.F. =NCERT Exemplar - Differential Equations Notes | EduRev
∴ Solution is
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Hence, the correct option is (a).

Q.72. Solution of the differential equationNCERT Exemplar - Differential Equations Notes | EduRev
(a) x (y + cos x) = sin x + c 
(b) x (y – cos x) = sin x + c 
(c) xy cos x = sin x + c 
(d) x (y + cos x) = cos x + c
Ans. (a)
Solution.

The given differential equation isNCERT Exemplar - Differential Equations Notes | EduRev
Since, it is a linear differential equation
NCERT Exemplar - Differential Equations Notes | EduRev
Integrating factor I.F. =NCERT Exemplar - Differential Equations Notes | EduRev
∴ Solution is y × I.F. =NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
⇒ yx + x cos x = sin x + c
⇒ x (y + cos x) = sin x + c
Hence, the correct option is (a).

Q.73. The general solution of the differential equation (ex + 1) ydy = (y + 1) exdx is:
(a) (y + 1) = k (ex + 1) 
(b) y + 1 = ex + 1 + k 
(c) y = log [k (y + 1) (ex + 1)] 
(d)NCERT Exemplar - Differential Equations Notes | EduRev
Ans. (c)
Solution.

The given differential equation is
(ex + 1) y dy = (y + 1) ex dx
NCERT Exemplar - Differential Equations Notes | EduRev
Integrating both sides, we get
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev

⇒ y – log |y + 1| = log |ex + 1| + log k

⇒ y = log |y + 1| + log |ex + 1| + log k

⇒ y = log |k (y + 1) (ex + 1)|
Hence, the correct option is (c).

Q.74. The solution of the differential equationNCERT Exemplar - Differential Equations Notes | EduRev
(a) y = e x -y- x2 e- y + c 
(b)NCERT Exemplar - Differential Equations Notes | EduRev
(c)NCERT Exemplar - Differential Equations Notes | EduRev
(d)NCERT Exemplar - Differential Equations Notes | EduRev
Ans. (b)
Solution.

 The given differential equation is
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Integrating both sides, we have
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Hence, the correct option is (b).

Q.75. The solution of the differential equationNCERT Exemplar - Differential Equations Notes | EduRev
(a) y (1 + x2) = c + tan–1
(b)NCERT Exemplar - Differential Equations Notes | EduRev
(c) y log (1 + x2) = c + tan–1
(d) y (1 + x2) = c + sin–1 x
Ans. (a)
Solution.

 The given differential equation is
NCERT Exemplar - Differential Equations Notes | EduRev
Since, it is a linear differential equation
NCERT Exemplar - Differential Equations Notes | EduRev
Integrating factor I.F. =NCERT Exemplar - Differential Equations Notes | EduRev
∴ Solution is  y × I.F. =NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Hence, the correct option is (a).

Q.76. Fill in the blanks
(i) The degree of the differential equationNCERT Exemplar - Differential Equations Notes | EduRevis _________.
(ii) The degree of the differential equationNCERT Exemplar - Differential Equations Notes | EduRevis _________.

(iii) The number of arbitrary constants in the general solution of a differential equation of order three is _________.

(iv)NCERT Exemplar - Differential Equations Notes | EduRevis an equation of the type _________.
(v) General solution of the differential equation of the typeNCERT Exemplar - Differential Equations Notes | EduRev
is given by _________.
(vi) The solution of the differential equationNCERT Exemplar - Differential Equations Notes | EduRevis _________.
(vii) The solution of (1 + x2)NCERT Exemplar - Differential Equations Notes | EduRevis _________.

(viii) The solution of the differential equation ydx + (x + xy)dy = 0 is ______.

(ix) General solution ofNCERT Exemplar - Differential Equations Notes | EduRevis  _________.

(x) The solution of differential equation coty dx = xdy is   _________.

(xi) The integrating factor ofNCERT Exemplar - Differential Equations Notes | EduRevis  _________.
Ans.
(i) The degree of the differential equationNCERT Exemplar - Differential Equations Notes | EduRevis not defined.
(ii) The given differential equation isNCERT Exemplar - Differential Equations Notes | EduRev
Squaring both sides, we get
NCERT Exemplar - Differential Equations Notes | EduRev
So, the degree of the equation is 2.
(iii) The number of arbitrary constants in the solution is 3.
(iv) The given differential equationNCERT Exemplar - Differential Equations Notes | EduRevis of the 

typeNCERT Exemplar - Differential Equations Notes | EduRev
(v) General solution of the differential equation of the type
NCERT Exemplar - Differential Equations Notes | EduRev
(vi) The given differential equation isNCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Since, it is linear differential equation
NCERT Exemplar - Differential Equations Notes | EduRev
Integrating factor I.F. =NCERT Exemplar - Differential Equations Notes | EduRev
∴ Solution is
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Hence, the solution is y =NCERT Exemplar - Differential Equations Notes | EduRev
(vii) The given differential equation is
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Since it is a linear differential equation
NCERT Exemplar - Differential Equations Notes | EduRev
Integrating factor I.F. =NCERT Exemplar - Differential Equations Notes | EduRev
∴ Solution isNCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Hence, the required solution is y =NCERT Exemplar - Differential Equations Notes | EduRev

(viii) The given differential equation is
ydx + (x + xy) dy = 0

⇒ (x + xy) dy = – y dx  ⇒ x (1 + y) dy = – y dx
NCERT Exemplar - Differential Equations Notes | EduRev
Integrating both sides, we get
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev

⇒ log y + y = – log  x + log c
⇒ log x  + log y + log ey = log c
⇒ log (xy . ey) = log c
∴ xy = c e–y
Hence, the required solution is xy = c e–y.
(ix) The given differential equation isNCERT Exemplar - Differential Equations Notes | EduRev
Since, it  a linear differential equation
∴ P = 1 and Q = sin x
Integrating factor I.F. =NCERT Exemplar - Differential Equations Notes | EduRev
∴ Solution is  y × I.F. =NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev...(1)
LetNCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
From eq. (1) we get
NCERT Exemplar - Differential Equations Notes | EduRev
Hence, the required solution is
NCERT Exemplar - Differential Equations Notes | EduRev
(x) The given differential equation is cot y dx = x dy
NCERT Exemplar - Differential Equations Notes | EduRev
Integrating both sides, we get
NCERT Exemplar - Differential Equations Notes | EduRev
⇒ log sec y – log x = log c
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
∴ x = C sec y
Hence, the required solution is  x = C sec y .
(xi) The given differential equation is
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
HereNCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Hence, the required I.F. =NCERT Exemplar - Differential Equations Notes | EduRev

Q.77. State True or False for the following:
(i) Integrating factor of the differential of the formNCERT Exemplar - Differential Equations Notes | EduRev

is given byNCERT Exemplar - Differential Equations Notes | EduRev
(ii) Solution of the differential equation of the typeNCERT Exemplar - Differential Equations Notes | EduRev
is given by x.I.F. =NCERT Exemplar - Differential Equations Notes | EduRev
(iii) Correct substitution for the solution of the differential equation 
of the typeNCERT Exemplar - Differential Equations Notes | EduRevwhere f (x, y) is a homogeneous function of zero

degree is y = vx.
(iv) Correct substitution for the solution of the differential equation 
of the typeNCERT Exemplar - Differential Equations Notes | EduRevwhere g (x, y) is a homogeneous function of the degree zero is x = vy.
(v) Number of arbitrary constants in the particular solution of a differential equation of order two is two. 
(vi) The differential equation representing the family of circles x2 + (y – a)2 = a2 will be of order two.
(vii) The solution ofNCERT Exemplar - Differential Equations Notes | EduRev
(viii) Differential equation representing the family of curves
NCERT Exemplar - Differential Equations Notes | EduRev
(ix) The solution of the differential equationNCERT Exemplar - Differential Equations Notes | EduRev
(x) Solution ofNCERT Exemplar - Differential Equations Notes | EduRev
(xi) The differential equation of all non horizontal lines in a plane is
NCERT Exemplar - Differential Equations Notes | EduRev
Ans.
(i) True
I.F. of the given differential equation
NCERT Exemplar - Differential Equations Notes | EduRev
(ii) True
(iii) True
(iv) True
(v) False
Since particular solution of a differential equation has no arbitrary constant.
(vi) False
We know that the order of the differential equation is equal to the number of arbitrary constants.
(vii) True The given differential equation is
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Integrating both sides, we get
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
(viii) True
Given equation is y = ex (A cos x + B sin x)
Differentiating both sides, we get
NCERT Exemplar - Differential Equations Notes | EduRev
Again differentiating w.r.t. x, we get
NCERT Exemplar - Differential Equations Notes | EduRev
(ix) True
The given differential equation is
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Here,NCERT Exemplar - Differential Equations Notes | EduRev
Integrating factor I.F. =NCERT Exemplar - Differential Equations Notes | EduRev
∴ Solution is y × I.F. =NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
⇒ y = – x + cx2 ⇒ y + x = cx2
(x) True The given differential equation is
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
Integrating both sides, we get
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
NCERT Exemplar - Differential Equations Notes | EduRev
(xi) True
Let y = mx + c be the non-horizontal line in a plane
NCERT Exemplar - Differential Equations Notes | EduRev

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