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**SHORT ANSWER TYPE QUESTIONS**

**Q.1. Find the solution of****Ans.**

The given differential equation is

Separating the variables, we get

Integrating both sides, we get

â‡’ â€“ 2^{â€“y} + 2^{â€“x} = c log 2

â‡’ 2^{â€“x} â€“ 2^{â€“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 that****and y = 0 when x = 5. ****Find the value of x when y = 3.****Ans.**

Given equation is

â‡’

Integrating both sides, we get

Put y = 0 and x = 5

âˆ´ The equation becomes

Now putting y = 3, we get

Hence the required value of x =

Q.4. Solve the differential equation (x^{2} â€“ 1)**Ans.**

Given differential equation is

Dividing by (x^{2} â€“ 1), we get

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

Integrating factor I.F. =

âˆ´ Solution of the equation is

Hence the required solution is **Q.5. Solve the differential equation****Ans.**

Given equation is

â‡’

Integrating both sides, we have

â‡’ log y = x â€“ x^{2} + log c

â‡’ log y â€“ log c = x â€“ x^{2}

â‡’

âˆ´

Hence, the required solution is**Q.6. Find the general solution of****Ans.**

Given equation is

Here, P = a and Q = e^{mx}

Solution of equation is y x I.F =

â‡’

â‡’

âˆ´

Hence the required solution is**Q.7. Solve the differential equation****Ans.**

Given that:

Put x + y = t

âˆ´

âˆ´

Integrating both sides, we have

â‡’

Hence, the required solution is ( x + c) . e ^{xy} + 1= 0.**Q.8. Solve: ydx â€“ xdy = x ^{2}ydx.**

Given equation is ydx â€“ xdy = x

â‡’ y dx - x

â‡’

â‡’

Integrating both sides we get

â‡’

âˆ´

Hence, the required solution is y =

Given equation is

â‡’

â‡’

Integrating both sides, we get

Put x = 0 and y = 0, we get tan

â‡’ c = 0

âˆ´

Hence, the required solution is

Given equation is

â‡’

â‡’

Here

âˆ´ Integrating factor I.F. =

So the solution of the equation is

â‡’

So x = y

Hence, the required solution is x = y (y

Given equation is

Integrating both sides, we get

â‡’

â‡’

Put x = 0 and y = 1, we get

(1 + 1) (2 + sin 0) = c â‡’ 4 = c

âˆ´ equation is (1 + y) (2 + sin x) = 4

Now put

âˆ´

â‡’

Hence, the required solution is

**y (1) =****Ans.**

Given equation is

Here,

âˆ´ Integrating factor I.F.

âˆ´ I.F. = e^{â€“t} . (1 + t)

Required solution of the given differential equation is

â‡’

â‡’

â‡’

Put t = 0 and y = â€“1 [âˆµ y(0) = â€“1]

â‡’

â‡’

So the equation becomes

ye^{â€“t} (1 + t) = â€“e^{â€“t}

Now put t = 1

âˆ´ y . e^{â€“1} (1 + 1) = â€“e^{â€“1}

â‡’

Hence y (1) =is verified.**Q.13. Form the differential equation having y = (sin ^{â€“1}x)^{2} + Acos^{â€“1}x + B, where A and B are arbitrary constants, as its general solution.**

Given equation is y = (sin

Multiplying both sides bywe get

Again differentiating w.r.t x, we get

â‡’

â‡’

Multiplying both sides by, we get

â‡’

Which is the required differential equation.

Equation of circle which passes through the origin and whose centre lies on y-axis is

(x â€“ 0)

â‡’ x

â‡’ x

Differentiating both sides w.r.t. x we get

â‡’

â‡’

âˆ´

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

â‡’

âˆ´

Hence, the required differential equation is

Given equation is

â‡’

Here,

Integrating factor I.F. =

âˆ´ Solution is

â‡’

â‡’

â‡’...(i)

Since the curve is passing through origin i.e., (0, 0)

âˆ´ Put y = 0 and x = 0 in eq. (i)

âˆ´ Equation is

Hence, the required solution is

Given equation is

â‡’

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

âˆ´

â‡’

Integrating both sides, we get

â‡’

Hence, the required solution is

Given equation is

â‡’

Here,

âˆ´ Integrating factor I.F. =

âˆ´ Solution is

â‡’

â‡’

Hence, this is the required general solution.

The given equation is y

â‡’ y

â‡’

Since it is a homogeneous differential equation

So,

â‡’

â‡’

â‡’

Integrating both sides, we get

â‡’

â‡’

Hence the required solution is

Given differential equation is (x + y) (dx â€“ dy) = dx + dy

â‡’ (x + y) dx â€“ (x + y) dy = dx + dy

â‡’ â€“ (x + y) dy â€“ dy = dx â€“ (x + y) dx

â‡’ â€“ (x + y + 1) dy = â€“ (x + y â€“ 1) dx

â‡’

Put x + y = z

âˆ´

So,

â‡’

â‡’

Integrating both sides, we get

â‡’

âˆ´ x + y = c . e

Hence, the required solution is x + y = c . e

Given differential equation is

â‡’

â‡’

Integrating both sides, we get

â‡’

â‡’

â‡’

Put x = 1, y = â€“2

â‡’

â‡’

â‡’ â€“ 1 â€“ 0 = 0 + c [âˆµ log (1) = 0]

âˆ´ c = â€“1

âˆ´ equation is

Hence, the required solution is x^{2} (y + 3)^{3} = e^{y + 2}**Q.21. Solve the differential equation dy = cos x (2 â€“ y cosec x) dx given that y = 2 when****Ans.**

The given differential equation is

dy = cos x (2 â€“ y cosec x) dx

Here, P = cot x and Q = 2 cos x.

âˆ´ Integrating factor I.F. =

âˆ´ Required solution is y Ã— I.F =

Put x =and y = 2, we get

â‡’

âˆ´ The equation is y sin x =**Q.22. Form the differential equation by eliminating A and B in Ax ^{2} + By^{2} = 1.**

Given that Ax

Differentiating w.r.t. x, we get

â‡’

âˆ´

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

Hence, the required equation is

xy.y â€³ + x.( y â€²)

Given differential equation is

(1 + y

â‡’ 2y (1 + x

â‡’

Integrating both sides, we get

â‡’

â‡’

Which is the required solution.

(x â€“ 1)

Differentiating both sides w.r.t., x we get

Which is the required equation.

**LONG ANSWER TYPE QUESTIONS**

**Q.25. Solve :****Ans.**

The given differential equation is

â‡’

â‡’

â‡’

Here, P =and Q = (sin x + log x)

Integrating factor I.F. =

âˆ´ Solution is

â‡’...(1)

Let I

Now from eq (1) we get,

Hence, the required solution is**Q.26. Find the general solution of (1 + tany) (dx â€“ dy) + 2xdy = 0.****Ans.**

Given that: (1 + tan y) (dx â€“ dy) + 2xdy = 0

â‡’ (1 + tan y) dx â€“ (1 + tan y) dy + 2xdy = 0

â‡’ (1 + tan y) dx â€“ (1 + tan y â€“ 2x) dy = 0

â‡’

Here, P =

Integrating factor I.F.

So, the solution is x Ã— I.F. =

â‡’ x . e^{y} (sin y + cos y) =

â‡’ x . e^{y} (sin y + cos y) = e^{y} . sin y + c

â‡’ x(sin y + cos y) = sin y + c . e^{â€“y}

Hence, the required solution is x(sin y + cos y) = sin y + c . e^{â€“y}.**Q.27. Solve : [Hint: Substitute x + y = z]****Ans.**

Given that :

Put x + y = v, on differentiating w.r.t. x, we get,

âˆ´

âˆ´-1 = cos v + sin v

â‡’

Integrating both sides, we have

â‡’

â‡’

Put

â‡’

â‡’

â‡’

â‡’

â‡’

Hence, the required solution is

= x + K [c â€“ log 2 = K]**Q.28. Find the general solution of ****Ans.**

Given equation is

Here, P = â€“3 and Q = sin 2 x

âˆ´ Integrating factor I.F. =

âˆ´ Solution is

â‡’

Let

â‡’

â‡’

â‡’

â‡’

â‡’

â‡’

â‡’

â‡’

âˆ´ The equation becomes

âˆ´

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

â‡’

â‡’

Integrating both sides, we get

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

âˆ´

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

â‡’[making perfect square]

â‡’

â‡’

â‡’

â‡’

âˆ´

Since, the line is passing through the point (1, 0), then

(0 â€“ 1) (1 + 1) = c(1) â‡’ c = 2.

Hence, the required solution is (y â€“ 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 â€“ y.

âˆ´ As per the condition,

Put x â€“ y = v

âˆ´

âˆ´ the equation becomes

Integrating both sides, we get

â‡’...(1)

Since, the curve is passing through (0, 0)

then

âˆ´ On putting c = 0 in eq. (1) we get

âˆ´

â‡’ (1 + x â€“ y) = e^{2x} (1 â€“ x + y)

Hence, the required equation is (1 + x â€“ y) = e^{2x} (1 â€“ 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)

âˆ´ Coordinates of A and B are (2x, 0) and (0, 2y) respectively.

âˆ´ Slope of the tangent

AB =

âˆ´

Integrating both sides, we get

â‡’ log y + log x = log c â‡’ log yx = log c

âˆ´ yx = c

Since, the curve passes through (1, 1)

âˆ´ 1 Ã— 1 = c âˆ´ c = 1

â‡’ yx = 1

Hence, the required equation is xy = 1.**Q.33. Solve : ****Ans.**

Given that:

Since, it is a homogeneous differential equation.

âˆ´ Put

â‡’

â‡’

Integrating both sides, we get

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

dv = dt

âˆ´

log|t| = log|x| + log c

â‡’ log |log v| = log xc â‡’ log v = xc

â‡’

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**

(d) 2

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).

Solution.

Given differential equation is

â‡’

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).

Solution.

Given equation is y = e

Differentiating both sides, w.r.t. x, we get

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

Hence, the correct option is (c)

Solution.

Given equation is : y = A cos Î± x + B sin Î± x

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

= â€“A sin Î± x . Î± + B cos Î± x . Î±

= â€“A Î± sin Î± x + B Î± cos Î± x

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

= â€“A Î±

â‡’= â€“ Î±

â‡’= â€“ Î±

Hence, the correct option is (b).

Solution.

The given differential equation is

xdy â€“ ydx = 0

â‡’

Integrating both sides, we get

â‡’ log y = log x + log c â‡’ log y = log xc

â‡’ 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 x ****(c) sec x ****(d) sin x****Ans. **(c)

Solution.

The given differential equation is

â‡’

Here, P = tan x and Q = sec x

âˆ´ Integrating factor =

Hence, the correct option is (c).**Q.41. Solution of the differential equation tany sec ^{2}x dx + tanx sec^{2}ydy = 0 is :**

Solution.

The given differential equation is

tan y sec

â‡’ tan x sec

â‡’

Integrating both sides, we get

Solution.

Given equation is y = Ax + A

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) â€“ x****Ans. **(c)

Solution.

The given differential equation is

â‡’

Here,

So, integrating factor =

Hence, the correct option is (c).**Q.44. Solution ofy (0) = 1 is given by****(a) xy = â€“ e ^{x} **

Solution.

The given differential equation is

Here, P = â€“1, Q = 1

âˆ´ Integrating factor, I.F. =

So, the solution is

â‡’

â‡’ y . e^{â€“x} = â€“ e^{â€“x} + c

Put x = 0, y = 1

â‡’ 1 . e^{0} = â€“ e^{0} + c

â‡’ 1 = â€“1 + c âˆ´ c = 2

So the equation is y . e^{â€“x} = â€“e^{â€“x} + 2

â‡’ y = â€“ 1 + 2e^{x} = 2e^{x} â€“ 1

Hence, the correct option is (d).**Q.45. The number of solutions of****when y (1) = 2 is**

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

**Ans. **(b)

Solution.

The given differential equation is

â‡’

Integrating both sides, we get

â‡’ log (y + 1) = log (x â€“ 1) + log c

â‡’ log (y + 1) â€“ log (x â€“ 1) = log c

â‡’

Put x = 1 and y = 2

â‡’âˆ´ c = âˆž

âˆ´â‡’ x â€“ 1 = 0 â‡’ x = 1

Hence, the correct option is (b).**Q.46. Which of the following is a second order differential equation?****(a) (yâ€²) ^{2} + x = y^{2} **

Solution.

Second order differential equation is yâ€²yâ€³+ y = sin x

Hence, the correct option is (b).

Solution.

The given differential equation is

â‡’

Here,

âˆ´ Integrating factor

Hence, the correct option is (c).

(d) (1 + x

Solution.

Given equation is tan

Differentiating w.r.t. x, we have

â‡’

â‡’(1 + x^{2}) dy = â€“ (1 + y^{2}) dx

â‡’ (1 + x^{2}) dy + (1 + y^{2}) 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

â‡’= c â€“ x â‡’ y dy = (c â€“ x) dx

âˆ´ Integrating both sides, we get

â‡’

â‡’ x^{2} + y^{2} â€“ 2cx = 2k which is a family of circles.

Hence, the correct option is (d).

**Q.50. The general solution of e ^{x} cosy dx â€“ e^{x} siny dy = 0 is :**

Solution.

The given differential equation is

e^{x} cos y dx â€“ e^{x} sin y dy = 0

â‡’ e^{x} (cos y dx â€“ sin y dy) = 0

â‡’ cos y dx - sin y dy = 0 [âˆµ e^{x} â‰ 0]

â‡’ sin y dy = cos y dx

Integrating both sides, we have

â‡’ - log |cos y| = x + log k â‡’

â‡’

â‡’

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 = e ^{x} (x â€“ 1) **

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 âˆ´ c = 0

So, the solution is y e

Hence, the correct option is (b).

Solution.

Given differential equation is

Here, P = tan x and Q = sec x

Hence, the correct option is (b).

Solution.

The given differential equation is

â‡’

Integrating both sides, we get

â‡’tan

â‡’

â‡’

â‡’ y â€“ x = k (1 + xy)

Hence, the correct option is (b).

(d) e

Solution.

The given differential equation is

â‡’

â‡’

Here, P =

âˆ´ Integrating factor I.F.

Hence, the correct option is (b).

Solution.

The given equation is y = ae

On differentiation, we get

Again differentiating w.r.t., we have

Hence, the correct option is (c).

Solution.

The given differential equation is

cos x sin y dx + sin x cos y dy = 0

â‡’ sin x cos y d y = â€“ cos x sin y dx

Integrating both sides, we have

â‡’

â‡’ log |sin y| = â€“ log |sin x| + log c

â‡’ log |sin y| + log |sin x| = log c

â‡’ log |sin y . sin x| = log c â‡’ sin x sin y = c

Hence, the correct option is (b).

(c) y = x . e

(d)

Solution.

The given differential equation is

â‡’

Here P

âˆ´ Integrating factor I.F. =

So, the solution is

Hence, the correct option is (a).

Solution.

The given equation is

x

Differentiating w.r.t. x, we have

â‡’

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

â‡’

âˆ´ Hence the correct option is (a).

Solution.

The given equation is

y = Ax + A

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

â‡’

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

âˆ´

Integrating both sides, we get

â‡’

â‡’ y^{2} â€“ x^{2} = 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 = â€“x and Q =

âˆ´ Integrating factor I.F. =

So, the solution is

â‡’

â‡’

â‡’

âˆ´

Hence the correct option is (c).**Q.64. The solution of the equation (2y â€“ 1) dx â€“ (2x + 3)dy = 0 is :****(a)****(b)****(c)****(d)****Ans.** (c)

Solution.

The given differential equation is

(2y â€“ 1) dx â€“ (2x + 3) dy = 0 â‡’ (2x + 3) dy = (2y - 1) dx

â‡’

Integrating both sides, we get

â‡’

â‡’log |2y â€“ 1| = log |2x + 3| + 2 log c

â‡’ log |2y â€“ 1| â€“ log |2x + 3| = log c^{2}

â‡’

â‡’

â‡’

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

â‡’

Hence, the correct option is (a).

**Q.66. The solution of****(a) y = e ^{â€“x} (x â€“ 1) **

Solution.

The given differential equation is

Since, it is a linear differential equation then P = 1 and Q = e

Integrating factor I.F. =

âˆ´ Solution is

Put y = 0 and x = 0

âˆ´ 0 = 0 + c âˆ´ c = 0

âˆ´ equation is y Ã— e^{x} = x

So y = x . e^{â€“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

âˆ´ the order of the differential equation is 3

and since, the power of highest order is 2

âˆ´ 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,

âˆ´ order = 2

and the power of the highest derivative is 1

âˆ´ degree = 1

Hence, the correct option is (c).**Q.69. The differential equation of the family of curves y ^{2} = 4a (x + a) is :**

Solution.

The given equation of family of curves is

y^{2} = 4a (x + a)

â‡’ y^{2} = 4ax + 4a^{2} ...(1)

Differentiating both sides, w.r.t. x, we get

â‡’

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) . e ^{x} **

Solution.

The given differential equation is

Since the above equation is of second order and first degree

âˆ´ D

â‡’ (D

âˆ´ auxiliary equation is

m

If the roots of Auxiliary equation are real and equal say (m)

then CF = (c

âˆ´ CF = (Ax + B) e

So y = (Ax + B) . e

Hence, the correct option is (a).

Solution.

The given differential equation is

Since, it is a linear differential equation

âˆ´ P = tan x and Q = sec x

Integrating factor I.F. =

âˆ´ Solution is

Hence, the correct option is (a).

Solution.

The given differential equation is

Since, it is a linear differential equation

Integrating factor I.F. =

âˆ´ Solution is y Ã— I.F. =

â‡’

â‡’

â‡’ yx + x cos x = sin x + c

â‡’ x (y + cos x) = sin x + c

Hence, the correct option is (a).

Solution.

The given differential equation is

(e

â‡’

Integrating both sides, we get

â‡’

â‡’

â‡’ y â€“ log |y + 1| = log |e^{x} + 1| + log k

â‡’ y = log |y + 1| + log |e^{x} + 1| + log k

â‡’ y = log |k (y + 1) (e^{x} + 1)|

Hence, the correct option is (c).**Q.74. The solution of the differential equation****(a) y = e ^{x -y}- x^{2} e^{- y} + c **

Solution.

The given differential equation is

â‡’

â‡’

Integrating both sides, we have

â‡’

Hence, the correct option is (b).

Solution.

The given differential equation is

Since, it is a linear differential equation

Integrating factor I.F. =

âˆ´ Solution is y Ã— I.F. =

â‡’

â‡’

Hence, the correct option is (a).

**(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 + x ^{2})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

â‡’

Since, it is linear differential equation

âˆ´

Integrating factor I.F. =

âˆ´ Solution is

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

â‡’

Since it is a linear differential equation

Integrating factor I.F. =

âˆ´ Solution is

â‡’

Hence, the required solution is y =

**(viii)** The given differential equation is

ydx + (x + xy) dy = 0

â‡’ (x + xy) dy = â€“ y dx â‡’ x (1 + y) dy = â€“ y dx

â‡’

Integrating both sides, we get

â‡’

â‡’ log y + y = â€“ log x + log c

â‡’ log x + log y + log e^{y} = log c

â‡’ log (xy . e^{y}) = log c

âˆ´ xy = c e^{â€“y}

Hence, the required solution is xy = c e^{â€“y}.**(ix)** The given differential equation is

Since, it a linear differential equation

âˆ´ P = 1 and Q = sin x

Integrating factor I.F. =

âˆ´ Solution is y Ã— I.F. =

â‡’...(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

â‡’ log sec y â€“ log x = log c

â‡’

âˆ´

âˆ´ x = C sec y

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

â‡’

â‡’

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 ****type****where 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 type****where 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 x ^{2} + (y â€“ a)^{2} = a^{2} will be of order two.**

I.F. of the given differential equation

Since particular solution of a differential equation has no arbitrary constant.

We know that the order of the differential equation is equal to the number of arbitrary constants.

â‡’

Integrating both sides, we get

â‡’

â‡’

Given equation is y = e

Differentiating both sides, we get

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

The given differential equation is

â‡’

Here,

Integrating factor I.F. =

âˆ´ Solution is y Ã— I.F. =

â‡’

â‡’

â‡’ y = â€“ x + cx

Integrating both sides, we get

âˆ´

Let y = mx + c be the non-horizontal line in a plane

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