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NCERT Exemplar - Differential Equations Notes | EduRev

JEE : NCERT Exemplar - Differential Equations Notes | EduRev

The document NCERT Exemplar - Differential Equations Notes | EduRev is a part of the JEE Course Mathematics (Maths) Class 12.
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Q.1. Find the solution of
Ans.
The given differential equation is

Separating the variables, we get

Integrating both sides, we get

⇒ – 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 get
Again differentiating w.r.t. x we have
Hence, the required equation is

Q.3. Given thatand y = 0 when x = 5. Find the value of x when y = 3.
Ans.
Given equation is

Integrating both sides, we get

Put y = 0 and x = 5

∴ The equation becomes
Now putting y = 3, we get

Hence the required value of x =

Q.4. Solve the differential equation (x2 – 1)

Ans.
Given differential equation is

Dividing by (x2 – 1), we get

It is a linear differential equation of first order and first degree.

Integrating factor I.F. =
∴ Solution of the equation is

Hence the required solution is

Q.5. Solve the differential equation
Ans.
Given equation is

Integrating both sides, we have

⇒ log y =  x – x2 + log c
⇒ log y – log c = x – x2

Hence, the required solution is

Q.6. Find the general solution of
Ans.
Given equation is
Here, P  = a and Q = emx

Solution of equation is y x I.F =

Hence the required solution is

Q.7. Solve the differential equation
Ans.
Given that:
Put x  + y = t

Integrating both sides, we have

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

Integrating both sides we get

Hence, the required solution is y =

Q.9. Solve the differential equation= 1 + x + y2 + xy2, when y = 0, x = 0.
Ans.
Given equation is

Integrating both sides, we get

Put x  = 0 and y = 0, we get tan–1(0) = 0 + 0 + c
⇒ c = 0

Hence, the required solution is

Q.10. Find the general solution of
Ans.
Given equation is

Here
∴ Integrating factor I.F. =
So the solution of the equation is

So x = y3 + cy = y(y2 + c)
Hence, the required solution is x = y (y2 + c).

Q.11. If y(x) is a solution of= – cos x and y (0) = 1, then find the value of
Ans.
Given equation is

Integrating both sides, we get

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 put

Hence, the required solution is

Q.12. If y(t) is a solution of (1 + t)and y (0) = – 1, then show that

y (1) =
Ans.
Given equation is

Here,
∴ Integrating factor I.F.

∴ I.F. = e–t . (1 + t)
Required solution of the given differential equation is

Put t = 0 and y = –1 [∵ y(0) = –1]

So the equation becomes
ye–t (1 + t) = –e–t
Now put t = 1
∴ y . e–1 (1 + 1) = –e–1

Hence y (1) =is 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

Multiplying both sides bywe get

Again differentiating w.r.t x, we get

Multiplying both sides by, we get

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

Putting the value of a in eq. (i), we get

Hence, the required differential equation is

Q.15. Find the equation of a curve passing through origin and satisfying the differential equation
Ans.
Given equation is

Here,
Integrating factor I.F. =
∴ Solution is

...(i)
Since the curve is passing through origin i.e., (0, 0)
∴ Put y = 0 and x = 0 in eq. (i)

∴ Equation is
Hence, the required solution is

Q.16. Solve :
Ans.
Given equation is

Put y  = vx  [∵ it is a homogeneous differential equation]

Integrating both sides, we get

Hence, the required solution is

Q.17. Find the general solution of the differential equation (1 + y2) +
(x – etan–1y)
Ans.
Given equation is

Here,
∴ Integrating factor I.F. =
∴ Solution is

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

Since it is a homogeneous differential equation

So,

Integrating both sides, we get

Hence the required solution is

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

Put x + y = z

So,

Integrating both sides, we get

∴ x + y = c . ex – y
Hence, the required solution is x + y = c . ex – y.

Q.20. Solve: 2(y + 3) – xygiven that y (1) = – 2.
Ans.
Given differential equation is

Integrating both sides, we get

Put x = 1, y = –2

⇒ – 1 – 0 = 0 + c [∵ log (1) = 0]
∴ c = –1
∴ equation is

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 when
Ans.
The given differential equation is
dy = cos x (2 – y cosec x) dx

Here, P = cot x and Q = 2 cos x.
∴ Integrating factor I.F. =
∴ Required solution is y × I.F =

Put x =and y = 2, we get

∴ The equation is y sin x =

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

Differentiating both sides again w.r.t. x, we have

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

Integrating both sides, we get

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

Which is the required equation.

Q.25. Solve :
Ans.
The given differential equation is

Here, P =and Q = (sin x + log x)
Integrating factor I.F. =
∴ Solution is

...(1)
Let I

Now from eq (1) we get,

Hence, the required solution is

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

Here, P =
Integrating factor I.F.

So, the solution is x × I.F. =
⇒ x . ey (sin y + cos y) =
⇒ x . ey (sin y + cos y) = ey . sin y + c

⇒ 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 : [Hint: Substitute x + y = z]
Ans.
Given that :
Put x + y = v, on differentiating w.r.t. x, we get,

-1 = cos v + sin v

Integrating both sides, we have

Put

Hence, the required solution is
= x + K [c – log 2 = K]

Q.28. Find the general solution of
Ans.
Given equation is
Here, P = –3 and Q = sin 2 x
∴ Integrating factor I.F. =
∴ Solution is

Let

∴ The equation becomes

Hence, the required solution is

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) is
Ans.
Given that the slope of tangent to a curve at (x, y) is

It is a homogeneous differential equation
So, put y = vx

Integrating both sides, we get

Since, the curve is passing through the point (2, 1)

Hence, the required equation is

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) is
Ans.
Given that the slope of the tangent to the curve at (x, y) is

Integrating both sides, we have

[making perfect square]

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 =
and the difference between the abscissa and ordinate = x – y.
∴ As per the condition,
Put x – y = v

∴ the equation becomes

Integrating both sides, we get

...(1)
Since, the curve is passing through (0, 0)
then
∴ On putting c = 0 in eq. (1) we get

⇒ (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

AB =

Integrating both sides, we get

⇒ 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 :
Ans.
Given that:

Since, it is a homogeneous differential equation.
∴ Put

Integrating both sides, we get

Put log v = t on L.H.S.
dv = dt

log|t| = log|x| + log c
⇒ log |log v| = log xc ⇒ log v = xc

Hence, the required solution is log

OBJECTIVE TYPE QUESTIONS

Q.34. The degree of the differential equation
(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 sinon expansion will be in increasing power of

Hence, the correct option is (d).

Q.35. The degree of the differential equation
(a) 4
(b) 3/2

(c) not defined
(d) 2

Ans. (d)
Solution.
The given differential equation is

Squaring both sides, we have

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 equation 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

Since the degree ofis 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)
(b)
(c)
(d)
Ans. (c)
Solution.

Given equation is y = e–x (A cos x + B sin x)
Differentiating both sides, w.r.t. x, we get

Again differentiating w.r.t. x, we get

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)
(b)
(c)
(d)
Ans. (b)
Solution.

Given equation is : y = A cos α x + B sin α x
Differentiating both sides w.r.t. x, we have
= –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
= –A α2 cos α x – B α2 sin α x
= – α2 (A cos α x + B sin α x)
= – α2 y ⇒
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

Integrating both sides, we get

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 cosx
(a) cos x
(b) tan
(c) sec
(d) sin x
Ans. (c)
Solution.

The given differential equation is

Here, P = tan x and Q = sec x
Integrating factor =
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)
(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

Integrating both sides, we get

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
= A which has degree 1.

Hence, the correct answer is (a).

Q.43. Integrating factor of

(a) x
(b) log x

(c) 1/x

(d) – x
Ans. (c)
Solution.

The given differential equation is

Here,
So, integrating factor =
Hence, the correct option is (c).

Q.44. Solution ofy (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

Here, P = –1, Q = 1
∴ Integrating factor, I.F. =
So, the solution is

⇒ 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 ofwhen y (1) = 2 is

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

Ans. (b)
Solution.

The given differential equation is

Integrating both sides, we get

⇒ log (y + 1) = log (x – 1) + log c
⇒ log (y + 1) – log (x – 1) = log c

Put x = 1 and y = 2
∴ c = ∞
⇒ 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)
(a) –x
(b)
(c)
(d)
Ans. (c)
Solution.

The given differential equation is

Here,
∴ Integrating factor

Hence, the correct option is (c).

Q.48. tan–1 x + tan–1 y = c is the general solution of the differential equation:
(a)
(b)
(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

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

Q.49. The differential equation yrepresents :
(a) Family of hyperbolas
(b) Family of parabolas
(c) Family of ellipses
(d) Family of circles
Ans. (d)
Solution.

Given differential equation is

= c – x ⇒ y dy = (c – x) dx
∴ Integrating both sides, we get

⇒ 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 dx
Integrating both sides, we have

⇒ - log |cos y| = x + log k ⇒

Hence, the correct option is (a).

Q.51. The degree of the differential equation    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 ofis
(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

Since, it is a linear differential equation

So, the solution is

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

(a) cos x
(b) sec x
(c) ecos x
(d) esec x
Ans. (b)
Solution.

Given differential equation is

Here, P = tan x and Q = sec x

Hence, the correct option is (b).

Q.54. The solution of the differential equation
(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

Integrating both sides, we get

⇒tan–1 y = tan–1 x + c ⇒ tan-1 y - tan-1 x = c

⇒ y – x = k (1 + xy)
Hence, the correct option is (b).

Q.55. The integrating factor of the differential equation
(a)
(b)
(c) xe
(d) ex

Ans. (b)
Solution.

The given differential equation is

Here, P =
∴ Integrating factor I.F.

Hence, the correct option is (b).

Q.56. y = aemx + be–mx satisfies which of the following differential equation?
(a)
(b)
(c)
(d)
Ans. (c)
Solution.

The given equation is  y = aemx + be–mx
On differentiation, we get
Again differentiating w.r.t., we have

Hence, the correct option is (c).

Q.57. The solution of the differential equation cosx siny dx + sinx cosy dy = 0
is :
(a)
(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

Integrating both sides, we have

⇒ 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 of
(a)
(b) y = xex + cx
(c) y = x . ex + k
(d)

Ans. (a)
Solution.

The given differential equation is

Here P
∴ Integrating factor I.F. =
So, the solution is

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)
(b)
(c)
(d)
Ans. (a)
Solution.

The given equation is
x2 + y2 – 2ay = 0 ..(1)
Differentiating w.r.t. x, we have

Putting the value of a in eq. (1) we get

∴ 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 get
Again differentiating both sides, we have

So the order of the differential equation is 2.

Hence, the correct option is (b).

Q.61. The general solution of

(a)
(b)
(c)
(d)
Ans. (c)
Solution.

The given differential equation is

Integrating both sides, we have

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

Integrating both sides, we get

⇒ y2 – x2 = 2c = k which is rectangular hyperbola.

Hence, the correct option is (d).

Q.63. The general solution of the differential equation

(a)
(b)
(c)
(d)
Ans. (c)
Solution.

The given differential equation is

Since it is linear differential equation where P = –x and Q =
∴ Integrating factor I.F. =
So, the solution is

Hence the correct option is (c).

Q.64. The solution of the equation (2y – 1) dx – (2x + 3)dy = 0 is :
(a)
(b)
(c)
(d)
Ans. (c)
Solution.

The given differential equation is
(2y – 1) dx – (2x + 3) dy = 0 ⇒ (2x + 3) dy = (2y - 1) dx

Integrating both sides, we get

⇒log |2y – 1| = log |2x + 3| + 2 log c
⇒ log |2y – 1| – log |2x + 3| = log c2

Hence, the correct option is (c).

Q.65. The differential equation for which y = acosx + bsinx is a solution, is :
(a)
(b)
(c)
(d)
Ans. (a)
Solution.

The given equation is
y = a cos x + b sin x

Hence, the correct option is (a).

Q.66. The solution of
(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 is
Since, it is a linear differential equation then P = 1 and Q = e–x
Integrating factor I.F. =
∴ Solution is

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

(a) 1, 4
(b) 3, 4
(c) 2, 4
(d) 3, 2
Ans. (d)
Solution.

The given differential equation is

Here the highest derivative is
∴ 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

(a)
(b) 2, 3
(c) 2, 1
(d) 3, 4
Ans. (c)
Solution.

The given differential equation is

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)
(b)
(c)
(d)
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

Now, putting the value of a in eq. (1) we get

Hence, the correct option is (d).

Q.70. Which of the following is the general solution of
(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

Since the above equation is of second order and first degree
∴ D2y – 2Dy + y = 0, where D =
⇒ (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 of
(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 is
Since, it is a linear differential equation
∴ P = tan x and Q = sec x
Integrating factor I.F. =
∴ Solution is

Hence, the correct option is (a).

Q.72. Solution of the differential equation
(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 is
Since, it is a linear differential equation

Integrating factor I.F. =
∴ Solution is y × I.F. =

⇒ 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)
Ans. (c)
Solution.

The given differential equation is
(ex + 1) y dy = (y + 1) ex dx

Integrating both sides, we get

⇒ 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 equation
(a) y = e x -y- x2 e- y + c
(b)
(c)
(d)
Ans. (b)
Solution.

The given differential equation is

Integrating both sides, we have

Hence, the correct option is (b).

Q.75. The solution of the differential equation
(a) y (1 + x2) = c + tan–1
(b)
(c) y log (1 + x2) = c + tan–1
(d) y (1 + x2) = c + sin–1 x
Ans. (a)
Solution.

The given differential equation is

Since, it is a linear differential equation

Integrating factor I.F. =
∴ Solution is  y × I.F. =

Hence, the correct option is (a).

Q.76. Fill in the blanks
(i) The degree of the differential equationis _________.
(ii) The degree of the differential equationis _________.

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

(iv)is an equation of the type _________.
(v) General solution of the differential equation of the type
is given by _________.
(vi) The solution of the differential equationis _________.
(vii) The solution of (1 + x2)is _________.

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

(ix) General solution ofis  _________.

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

(xi) The integrating factor ofis  _________.
Ans.
(i) The degree of the differential equationis not defined.
(ii) The given differential equation is
Squaring both sides, we get

So, the degree of the equation is 2.
(iii) The number of arbitrary constants in the solution is 3.
(iv) The given differential equationis of the

type
(v) General solution of the differential equation of the type

(vi) The given differential equation is

Since, it is linear differential equation

Integrating factor I.F. =
∴ Solution is

Hence, the solution is y =
(vii) The given differential equation is

Since it is a linear differential equation

Integrating factor I.F. =
∴ Solution is

Hence, the required solution is y =

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

⇒ (x + xy) dy = – y dx  ⇒ x (1 + y) dy = – y dx

Integrating both sides, we get

⇒ 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 is
Since, it  a linear differential equation
∴ P = 1 and Q = sin x
Integrating factor I.F. =
∴ Solution is  y × I.F. =
...(1)
Let

From eq. (1) we get

Hence, the required solution is

(x) The given differential equation is cot y dx = x dy

Integrating both sides, we get

⇒ log sec y – log x = log c

∴ x = C sec y
Hence, the required solution is  x = C sec y .
(xi) The given differential equation is

Here

Hence, the required I.F. =

Q.77. State True or False for the following:
(i) Integrating factor of the differential of the form

is given by
(ii) Solution of the differential equation of the type
is given by x.I.F. =
(iii) Correct substitution for the solution of the differential equation
of the typewhere 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 typewhere 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 of
(viii) Differential equation representing the family of curves

(ix) The solution of the differential equation
(x) Solution of
(xi) The differential equation of all non horizontal lines in a plane is

Ans.
(i) True
I.F. of the given differential equation

(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

Integrating both sides, we get

(viii) True
Given equation is y = ex (A cos x + B sin x)
Differentiating both sides, we get

Again differentiating w.r.t. x, we get

(ix) True
The given differential equation is

Here,
Integrating factor I.F. =
∴ Solution is y × I.F. =

⇒ y = – x + cx2 ⇒ y + x = cx2
(x) True The given differential equation is

Integrating both sides, we get

(xi) True
Let y = mx + c be the non-horizontal line in a plane

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Mathematics (Maths) Class 12

209 videos|222 docs|124 tests

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