CBSE Class 12 Mathematics Chapter 9 Differential Equations – NCERT Solutions 2025-26

Exercise 9.5

1. Solve the differential equation \[\dfrac{{dy}}{{dx}} + 2y = \sin x\].

Ans: The given equation is in the form of \[\dfrac{{dy}}{{dx}} + Py = Q\]

Where, \[P = 2\] and \[Q = \sin x\]

Calculating integration factor,

\[I.F. = {e^{\int {Pdx} }}\]

\[I.F. = {e^{\int {2dx} }}\]

\[I.F. = {e^{2x}}\]

Therefore, solution of the given differential equation is given by the relation,

\[y\left( {I.F.} \right) = \int {\left( {Q \times I.F.} \right)dx}  + C\]

\[y{e^{2x}} = \int {\sin x.{e^{2x}}dx}  + C.......\left( 1 \right)\]

Let \[I = \int {\sin x.{e^{2x}}dx} \]

Using integration by parts,

\[I = \sin x\int {{e^{2x}}dx}  - \int {\left( {\dfrac{{d\left( {\sin x} \right)}}{{dx}}.\int {{e^{2x}}dx} } \right)} dx\]

\[I = \dfrac{{{e^{2x}}\sin x}}{2} - \int {\cos x.\dfrac{{{e^{2x}}}}{2}dx} \]

Again using integration by parts,

\[I = \dfrac{{{e^{2x}}\sin x}}{2} - \cos x\int {\dfrac{{{e^{2x}}}}{2}dx}  + \int {\left( {\dfrac{{d\left( {\cos x} \right)}}{{dx}}.\int {\dfrac{{{e^{2x}}}}{2}dx} } \right)} dx\]

\[I = \dfrac{{{e^{2x}}\sin x}}{2} - \dfrac{{{e^{2x}}\cos x}}{4} + \int {\left( { - \sin x} \right).\dfrac{{{e^{2x}}}}{4}dx} \]

\[I = \dfrac{{{e^{2x}}\sin x}}{2} - \dfrac{{{e^{2x}}\cos x}}{4} - \dfrac{1}{4}\int {\sin x.{e^{2x}}dx} \]

As \[I = \int {\sin x.{e^{2x}}dx} \],

Therefore,

\[I = \dfrac{{{e^{2x}}\sin x}}{2} - \dfrac{{{e^{2x}}\cos x}}{4} - \dfrac{1}{4}I\]

\[I + \dfrac{1}{4}I = \dfrac{{{e^{2x}}\sin x}}{2} - \dfrac{{{e^{2x}}\cos x}}{4}\]

\[\dfrac{5}{4}I = \dfrac{{{e^{2x}}\sin x}}{2} - \dfrac{{{e^{2x}}\cos x}}{4}\]

\[I = \dfrac{{{e^{2x}}}}{5}\left( {2\sin x - \cos x} \right)\]

Substituting this value of I in equation 1.

\[y{e^{2x}} = \dfrac{{{e^{2x}}}}{5}\left( {2\sin x - \cos x} \right) + C\]

\[y = \dfrac{1}{5}\left( {2\sin x - \cos x} \right) + C{e^{ - 2x}}\]

This required differential equation .

2. Solve the differential equation \[\dfrac{{dy}}{{dx}} + 3y = {e^{ - 2x}}\].

Ans: The given equation is in the form of \[\dfrac{{dy}}{{dx}} + Py = Q\]

Where, \[P = 3\] and \[Q = {e^{ - 2x}}\]

Calculating integration factor,

\[I.F. = {e^{\int {Pdx} }}\]

\[I.F. = {e^{\int {3dx} }}\]

\[I.F. = {e^{3x}}\]

Therefore, solution of the given differential equation is given by the relation,

\[y\left( {I.F.} \right) = \int {\left( {Q \times I.F.} \right)dx}  + C\]

\[y{e^{3x}} = \int {\left( {{e^{3x}} \times {e^{ - 2x}}} \right)dx}  + C\]

\[y{e^{3x}} = \int {{e^x}dx}  + C\]

\[y{e^{3x}} = {e^x} + C\]

\[y = {e^{ - 2x}} + C{e^x}\]

This required differential equation.

3. Solve the differential equation \[\dfrac{{dy}}{{dx}} + \dfrac{y}{x} = {x^2}\].

Ans: The given equation is in the form of \[\dfrac{{dy}}{{dx}} + Py = Q\]

Where, \[P = \dfrac{1}{x}\] and \[Q = {x^2}\]

Calculating integration factor,

\[I.F. = {e^{\int {Pdx} }}\]

\[I.F. = {e^{\int {\dfrac{1}{x}dx} }}\]

\[I.F. = {e^{\log x}}\]

\[I.F. = x\]

Therefore, solution of the given differential equation is given by the relation,

\[y\left( {I.F.} \right) = \int {\left( {Q \times I.F.} \right)dx}  + C\]

\[yx = \int {\left( {{x^2}.x} \right)dx}  + C\]

\[yx = \int {\left( {{x^3}} \right)dx}  + C\]

\[yx = \dfrac{{{x^4}}}{4} + C\]

This required differential equation.

4. Solve the differential equation \[\dfrac{{dy}}{{dx}} + \left( {\sec x} \right)y = \tan x{\text{ }}\left( {0 \leqslant x \leqslant \dfrac{\pi }{2}} \right)\]

Ans: The given equation is in the form of \[\dfrac{{dy}}{{dx}} + Py = Q\]

Where, \[P = \sec x\] and \[Q = \tan x\]

Calculating integration factor,

\[I.F. = {e^{\int {Pdx} }}\]

\[I.F. = {e^{\int {\sec xdx} }}\]

\[I.F. = {e^{\log \left( {\sec x + \tan x} \right)}}\]

\[I.F. = \sec x + \tan x\]

Therefore, solution of the given differential equation is given by the relation,

\[y\left( {I.F.} \right) = \int {\left( {Q \times I.F.} \right)dx}  + C\]

\[y\left( {\sec x + \tan x} \right) = \int {\tan x\left( {\sec x + \tan x} \right)dx}  + C\]

\[y\left( {\sec x + \tan x} \right) = \int {\sec x\tan xdx}  + \int {{{\tan }^2}xdx}  + C\]

\[y\left( {\sec x + \tan x} \right) = \sec x + \int {\left( {{{\sec }^2}x - 1} \right)dx}  + C\]

\[y\left( {\sec x + \tan x} \right) = \sec x + \tan x - x + C\]

This required differential equation.

5. Solve the differential equation \[{\cos ^2}x\dfrac{{dy}}{{dx}} + y = \tan x{\text{ }}\left( {0 \leqslant x\dfrac{\pi }{2}} \right)\]

Ans: On rearranging the given equation,

\[\dfrac{{dy}}{{dx}} + {\sec ^2}x.y = {\sec ^2}x\tan x\]

The given equation is in the form of \[\dfrac{{dy}}{{dx}} + Py = Q\]

Where, \[P = {\sec ^2}x\] and \[Q = {\sec ^2}x\tan x\]

Calculating integration factor,

\[I.F. = {e^{\int {Pdx} }}\]

\[I.F. = {e^{\int {{{\sec }^2}xdx} }}\]

\[I.F. = {e^{\tan x}}\]

Therefore, solution of the given differential equation is given by the relation,

\[y\left( {I.F.} \right) = \int {\left( {Q \times I.F.} \right)dx}  + C\]

\[y{e^{\tan x}} = \int {\tan x{{\sec }^2}x{e^{\tan x}}dx + C} \]

Put \[\tan x = t\]

Differentiating w.r.t. \[t\].

\[{\sec ^2}xdx = dt\]

Therefore, above equation become,

\[y{e^{\tan x}} = \int {t{e^t}dt + C} \]

Using integration by parts,

\[y{e^{\tan x}} = t{e^t} - \int {{e^t}dt + C} \]

\[y{e^{\tan x}} = t{e^t} - {e^t} + C\]

Substituting the value of \[t\] .

\[y{e^{\tan x}} = \tan x{e^{\tan x}} - {e^{\tan x}} + C\]

\[y = \tan x - 1 + C{e^{ - \tan x}}\]

This is required differential equation.

6. Solve the differential equation \[x\dfrac{{dy}}{{dx}} + 2y = {x^2}\log x\]

Ans: On rearranging given equation can be written as,

\[\dfrac{{dy}}{{dx}} + \dfrac{2}{x}y = x\log x\]

The given equation is in the form of \[\dfrac{{dy}}{{dx}} + Py = Q\]

Where, \[P = \dfrac{2}{x}\] and \[Q = x\log x\]

Calculating integration factor,

\[I.F. = {e^{\int {Pdx} }}\]

\[I.F. = {e^{\int {\dfrac{2}{x}dx} }}\]

\[I.F. = {e^{\log {x^2}}}\]

\[I.F. = {x^2}\]

Therefore, solution of the given differential equation is given by the relation,

\[y\left( {I.F.} \right) = \int {\left( {Q \times I.F.} \right)dx}  + C\]

\[y{x^2} = \int {x\log x.{x^2}dx}  + C\]

\[y{x^2} = \int {{x^3}\log xdx}  + C\]

Using integration by parts,

\[y{x^2} = \log x\int {{x^3}dx}  - \int {\left( {\dfrac{d}{{dx}}\left( {\log x} \right)\int {{x^3}dx} } \right)dx}  + C\]

\[y{x^2} = \dfrac{{{x^4}\log x}}{4} - \int {\left( {\dfrac{1}{x}.\dfrac{{{x^4}}}{4}} \right)dx + C} \]

\[y{x^2} = \dfrac{{{x^4}\log x}}{4} - \dfrac{1}{4}.\dfrac{{{x^4}}}{4} + C\]

\[y{x^2} = \dfrac{{{x^4}\log x}}{4} - \dfrac{{{x^4}}}{{16}} + C\]

\[y = \dfrac{{{x^2}\log x}}{4} - \dfrac{{{x^2}}}{{16}} + C{x^{ - 2}}\]

\[y = \dfrac{{{x^2}}}{{16}}\left( {4\log x - 1} \right) + C{x^{ - 2}}\]

This is required differential equation.

7. Solve the differential equation \[x\log x\dfrac{{dy}}{{dx}} + y = \dfrac{2}{x}\log x\]

Ans: On rearranging the given equation can be written as,

\[\dfrac{{dy}}{{dx}} + \dfrac{y}{{\log x}} = \dfrac{2}{{{x^2}}}\]

The given equation is in the form of \[\dfrac{{dy}}{{dx}} + Py = Q\]

Where, \[P = \dfrac{1}{{x\log x}}\] and \[Q = \dfrac{2}{{{x^2}}}\]

Calculating integration factor,

\[I.F. = {e^{\int {Pdx} }}\]

\[\]

\[I.F. = {e^{\log \left( {\log x} \right)}}\]

\[I.F. = \log x\]

Therefore, solution of the given differential equation is given by the relation,

\[y\left( {I.F.} \right) = \int {\left( {Q \times I.F.} \right)dx}  + C\]

\[y\log x = \int {\left( {\dfrac{2}{{{x^2}}}\log x} \right)dx}  + C\]

Using integration by parts,

\[y\log x = 2\left[ {\log x\int {\dfrac{1}{{{x^2}}}dx}  - \int {\left( {\dfrac{d}{{dx}}\left( {\log x} \right)\int {\dfrac{1}{{{x^2}}}dx} } \right)dx} } \right] + C\]

\[y\log x = 2\left[ {\log x\left( { - \dfrac{1}{x}} \right) - \int {\left( {\dfrac{1}{x}\left( { - \dfrac{1}{x}} \right)} \right)dx} } \right] + C\]

\[y\log x = 2\left[ {\log x\left( { - \dfrac{1}{x}} \right) + \int {\dfrac{1}{{{x^2}}}dx} } \right] + C\]

\[y\log x = 2\left[ {\log x\left( { - \dfrac{1}{x}} \right) - \dfrac{1}{x}} \right] + C\]

\[y\log x =  - \dfrac{2}{x}\left( {1 + \log x} \right) + C\]

This is required differential equation.

8. Solve the differential equation \[\left( {1 + {x^2}} \right)dy + 2xydx = \cot xdx\]

Ans: On rearranging the given equation can be written as,

\[\dfrac{{dy}}{{dx}} + \dfrac{{2xy}}{{1 + {x^2}}} = \dfrac{{\cot x}}{{1 + {x^2}}}\]

The given equation is in the form of \[\dfrac{{dy}}{{dx}} + Py = Q\]

Where, \[P = \dfrac{{2x}}{{1 + {x^2}}}\] and \[Q = \dfrac{{\cot x}}{{1 + {x^2}}}\]

Calculating integration factor,

\[I.F. = {e^{\int {Pdx} }}\]

\[I.F. = {e^{\int {\dfrac{{2x}}{{1 + {x^2}}}dx} }}\]

\[I.F. = {e^{\log \left( {1 + {x^2}} \right)}}\]

\[I.F. = 1 + {x^2}\]

Therefore, solution of the given differential equation is given by the relation,

\[y\left( {I.F.} \right) = \int {\left( {Q \times I.F.} \right)dx}  + C\]

\[y\left( {1 + {x^2}} \right) = \int {\dfrac{{\cot x}}{{1 + {x^2}}} \times \left( {1 + {x^2}} \right)dx}  + C\]

\[y\left( {1 + {x^2}} \right) = \int {\cot xdx}  + C\]

\[y\left( {1 + {x^2}} \right) = \log \left( {\sin x} \right) + C\]

This is required differential equation.

9. Solve the differential equation \[x\dfrac{{dy}}{{dx}} + y - x + xy\cot x = 0\]

Ans: On rearranging given equation can be written as,

\[\dfrac{{dy}}{{dx}} + \left( {\dfrac{1}{x} + \cot x} \right)y = 1\]

The given equation is in the form of \[\dfrac{{dy}}{{dx}} + Py = Q\]

Where, \[P = \dfrac{1}{x} + \cot x\] and \[Q = 1\]

Calculating integration factor,

\[I.F. = {e^{\int {Pdx} }}\]

\[I.F. = {e^{\int {\left( {\dfrac{1}{x} + \cot x} \right)dx} }}\]

\[I.F. = {e^{\log x + \log \sin x}}\]

\[I.F. = x\sin x\]

Therefore, solution of the given differential equation is given by the relation,

\[y\left( {I.F.} \right) = \int {\left( {Q \times I.F.} \right)dx}  + C\]

\[yx\sin x = \int {\left( {1.x\sin x} \right)dx}  + C\]

\[yx\sin x = x\int {\sin xdx - \int {\left( {\dfrac{d}{{dx}}\left( x \right)\int {\sin xdx} } \right)dx} }  + C\]

\[yx\sin x = x\left( { - \cos x} \right) - \int {\left( { - \cos x} \right)dx}  + C\]

\[yx\sin x =  - x\cos x + \sin x + C\]

\[y = \dfrac{{ - x\cos x}}{{x\sin x}} + \dfrac{{\sin x}}{{x\sin x}} + \dfrac{C}{{x\sin x}}\]

\[y =  - \cot x + \dfrac{1}{x} + \dfrac{C}{{x\sin x}}\]

This is required differential equation.

10. Solve the differential equation \[\left( {x + y} \right)\dfrac{{dy}}{{dx}} = 1\]

Ans: On rearranging the given equation can be written as,

\[\dfrac{{dx}}{{dy}} - x = y\]

The given equation is in the form of \[\dfrac{{dx}}{{dy}} + Px = Q\]

Where, \[P =  - 1\] and \[Q = y\]

Calculating integration factor,

\[I.F. = {e^{\int { - 1dy} }}\]

\[I.F. = {e^{ - y}}\]

Therefore, solution of the given differential equation is given by the relation,

\[x\left( {I.F.} \right) = \int {\left( {Q \times I.F.} \right)dy}  + C\]

\[x{e^{ - y}} = \int {\left( {y{e^{ - y}}} \right)dy}  + C\]

\[x{e^{ - y}} = y\int {{e^{ - y}}dy}  - \int {\left( {\dfrac{d}{{dy}}\left( y \right)\int {{e^{ - y}}dy} } \right)dy}  + C\]

\[x{e^{ - y}} =  - y{e^{ - y}} - \int {\left( { - {e^{ - y}}} \right)dy}  + C\]

\[x{e^{ - y}} =  - y{e^{ - y}} - {e^{ - y}} + C\]

\[x =  - y - 1 + C{e^y}\]

\[x + y + 1 = C{e^y}\]

This is required differential equation.

11. Solve the differential equation \[ydx + \left( {x - {y^2}} \right)dy = 0\]

Ans: On rearranging the given equation can be written as,

\[\dfrac{{dx}}{{dy}} + \dfrac{x}{y} = y\]

The given equation is in the form of \[\dfrac{{dx}}{{dy}} + Px = Q\]

Where, \[P = \dfrac{1}{y}\] and \[Q = y\]

Calculating integration factor,

\[I.F. = {e^{\int {Pdy} }}\]

\[I.F. = {e^{\int {\dfrac{1}{y}dy} }}\]

\[I.F. = {e^{\log y}}\]

\[I.F. = y\]

Therefore, solution of the given differential equation is given by the relation,

\[x\left( {I.F.} \right) = \int {\left( {Q \times I.F.} \right)dy}  + C\]

\[xy = \int {\left( {y.y} \right)dy}  + C\]

\[xy = \int {{y^2}dy}  + C\]

\[xy = \dfrac{{{y^3}}}{3} + C\]

This is required differential equation.

12. Solve the differential equation \[\left( {x + 3{y^2}} \right)\dfrac{{dy}}{{dx}} = y{\text{ }}\left( {y > 0} \right)\]

Ans: On rearranging the given equation can be written as,

\[\dfrac{{dx}}{{dy}} - \dfrac{x}{y} = 3y\]

The given equation is in the form of \[\dfrac{{dx}}{{dy}} + Px = Q\]

Where, \[P =  - \dfrac{1}{y}\] and \[Q = 3y\]

Calculating integration factor,

\[I.F. = {e^{\int {Pdy} }}\]

\[I.F. = {e^{\int {\left( { - \dfrac{1}{y}} \right)dy} }}\]

\[I.F. = {e^{\log \left( {\dfrac{1}{y}} \right)}}\]

\[I.F. = \dfrac{1}{y}\]

Therefore, solution of the given differential equation is given by the relation,

\[x\left( {I.F.} \right) = \int {\left( {Q \times I.F.} \right)dy}  + C\]

\[x\left( {\dfrac{1}{y}} \right) = \int {\left( {3y \times \dfrac{1}{y}} \right)dy}  + C\]

\[\dfrac{x}{y} = \int {3dy}  + C\]

\[\dfrac{x}{y} = 3y + C\]

\[x = 3{y^2} + Cy\]

This is required differential equation.

13. Solve the differential equation \[\dfrac{{dy}}{{dx}} + 2y\tan x = \sin x;{\text{   }}y = 0{\text{ when }}x = \dfrac{\pi }{3}\]

Ans: The given equation is in the form of \[\dfrac{{dy}}{{dx}} + Py = Q\]

Where, \[P = 2\tan x\] and \[Q = \sin x\]

Calculating integration factor,

\[I.F. = {e^{\int {Pdx} }}\]

\[I.F. = {e^{\int {2\tan xdx} }}\]

\[I.F. = {e^{\log \left( {{{\sec }^2}x} \right)}}\]

\[I.F. = {\sec ^2}x\]

Therefore, solution of the given differential equation is given by the relation,

\[y\left( {I.F.} \right) = \int {\left( {Q \times I.F.} \right)dx}  + C\]

\[y{\sec ^2}x = \int {\left( {\sin x.{{\sec }^2}x} \right)dx}  + C\]

\[y{\sec ^2}x = \int {\left( {\sec x\tan x} \right)dx}  + C\]

\[y{\sec ^2}x = \sec x + C......\left( 1 \right)\]

Now, \[y = 0{\text{ when }}x = \dfrac{\pi }{3}\]

\[0 \times {\sec ^2}\left( {\dfrac{\pi }{3}} \right) = \sec \left( {\dfrac{\pi }{3}} \right) + C\]

\[C =  - 2\]

Substituting the value of \[C =  - 2\]in equation \[\left( 1 \right)\].

\[y{\sec ^2}x = \sec x - 2\]

\[y = \cos x - 2{\sec ^2}x\]

This is required differential equation.

14. Solve the differential equation \[\left( {1 + {x^2}} \right)\dfrac{{dy}}{{dx}} + 2xy = \dfrac{1}{{1 + {x^2}}}{\text{   }}y = 0{\text{ when }}x = 1\]

Ans: On rearranging the given equation can be written as,

\[\dfrac{{dy}}{{dx}} + \dfrac{{2xy}}{{1 + {x^2}}} = \dfrac{1}{{{{\left( {1 + {x^2}} \right)}^2}}}\]

The given equation is in the form of \[\dfrac{{dy}}{{dx}} + Py = Q\]

Where, \[P = \dfrac{{2x}}{{1 + {x^2}}}\] and \[Q = \dfrac{1}{{{{\left( {1 + {x^2}} \right)}^2}}}\]

Calculating integration factor,

\[I.F. = {e^{\int {Pdx} }}\]

\[I.F. = {e^{\int {\dfrac{{2x}}{{1 + {x^2}}}dx} }}\]

\[I.F. = {e^{\log \left( {1 + {x^2}} \right)}}\]

\[I.F. = 1 + {x^2}\]

Therefore, solution of the given differential equation is given by the relation,

\[y\left( {I.F.} \right) = \int {\left( {Q \times I.F.} \right)dx}  + C\]

\[y\left( {1 + {x^2}} \right) = \int {\left( {\dfrac{1}{{{{\left( {1 + {x^2}} \right)}^2}}}.\left( {1 + {x^2}} \right)} \right)dx}  + C\]

\[y\left( {1 + {x^2}} \right) = \int {\dfrac{1}{{1 + {x^2}}}dx}  + C\]

\[y\left( {1 + {x^2}} \right) = {\tan ^{ - 1}}x + C......\left( 1 \right)\]

Now, \[y = 0{\text{ when }}x = 1\]

\[0 = {\tan ^{ - 1}}1 + C\]

\[C =  - \dfrac{\pi }{4}\]

Substituting the value of \[C =  - \dfrac{\pi }{4}\] in equation\[\left( 1 \right)\],

\[y\left( {1 + {x^2}} \right) = {\tan ^{ - 1}}x - \dfrac{\pi }{4}\]

This is required differential equation.

15. Solve the differential equation \[\dfrac{{dy}}{{dx}} - 3y\cot x = \sin 2x{\text{        }}y = 2{\text{ when }}x = \dfrac{\pi }{2}\]

Ans: The given equation is in the form of \[\dfrac{{dy}}{{dx}} + Py = Q\]

Where, \[P =  - 3\cot x\] and \[Q = \sin 2x\]

Calculating integration factor,

\[I.F. = {e^{\int {Pdx} }}\]

\[I.F. = {e^{\int { - 3\cot xdx} }}\]

\[I.F. = {e^{ - 3\log \left( {\sin x} \right)}}\]

\[I.F. = \dfrac{1}{{{{\sin }^3}x}}\]

Therefore, solution of the given differential equation is given by the relation,

\[y\left( {I.F.} \right) = \int {\left( {Q \times I.F.} \right)dx}  + C\]

\[y.\dfrac{1}{{{{\sin }^3}x}} = \int {\left( {\sin 2x.\dfrac{1}{{{{\sin }^3}x}}} \right)dx}  + C\]

\[y.\cos e{c^3}x = 2\int {\left( {\cot x\cos ecx} \right)dx}  + C\]

\[y.\cos e{c^3}x =  - 2\cos ecx + C\]

\[y =  - \dfrac{2}{{\cos e{c^2}x}} + \dfrac{C}{{\cos e{c^3}x}}\]

\[y =  - 2{\sin ^2}x + C{\sin ^3}x......\left( 1 \right)\]

Now, \[y = 2{\text{ when }}x = \dfrac{\pi }{2}\]

\[2 =  - 2{\sin ^2}\left( {\dfrac{\pi }{2}} \right) + C{\sin ^3}\left( {\dfrac{\pi }{2}} \right)\]

\[2 =  - 2 + C\]

\[C = 4\]

Substituting the value of \[C = 4\]in equation 1,

\[y =  - 2{\sin ^2}x + 4{\sin ^3}x\]

\[y = 4{\sin ^3}x - 2{\sin ^2}x\]

This is required differential equation.

16. Find the equation of a curve passing through the origin given that the slope of thetangent to the curve at any point \[(x,y)\] is equal to the sum of the coordinates ofthe point.

Ans: Let \[f\left( {x,y} \right)\] be the curve passing through the origin.

At point \[(x,y)\], the slope of curve will be \[\dfrac{{dy}}{{dx}}\] .

According to the question,

\[\dfrac{{dy}}{{dx}} - y = x\]

The given equation is in the form of \[\dfrac{{dx}}{{dy}} + Px = Q\]

Where, \[P =  - 1\] and \[Q = x\]

Calculating integration factor,

\[I.F. = {e^{\int { - 1dx} }}\]

\[I.F. = {e^{ - x}}\]

Therefore, solution of the given differential equation is given by the relation,

\[x\left( {I.F.} \right) = \int {\left( {Q \times I.F.} \right)dy}  + C\]

\[y{e^{ - x}} = \int {\left( {x{e^{ - x}}} \right)dx}  + C\]

\[y{e^{ - x}} = x\int {{e^{ - x}}dy}  - \int {\left( {\dfrac{d}{{dx}}\left( x \right)\int {{e^{ - x}}dx} } \right)dx}  + C\]

\[y{e^{ - x}} =  - x{e^{ - x}} - \int {\left( { - {e^{ - x}}} \right)dx}  + C\]

\[y{e^{ - x}} =  - x{e^{ - x}} - {e^{ - x}} + C\]

\[y =  - x - 1 + C{e^x}\]

\[x + y + 1 = C{e^x}......\left( 1 \right)\]

As, equation is passing through the origin.

Therefore,

\[0 + 0 + 1 = C{e^0}\]

\[1 = C\]

Substituting the value of \[1 = C\] in equation 1.

\[x + y + 1 = {e^x}\]

This is required differential equation of curve passing through origin.

17. Find the equation of a curve passing through the point \[\left( {0,2} \right)\] given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope ofthe tangent to the curve at that point by 5.

Ans: Let \[f\left( {x,y} \right)\] be the curve passing through the origin.

At point \[(x,y)\], the slope of curve will be \[\dfrac{{dy}}{{dx}}\] .

According to the question,

\[\dfrac{{dy}}{{dx}} + 5 = x + y\]

\[\dfrac{{dy}}{{dx}} - y = x - 5\]

The given equation is in the form of \[\dfrac{{dx}}{{dy}} + Px = Q\]

Where, \[P =  - 1\] and \[Q = x - 5\]

Calculating integration factor,

\[I.F. = {e^{\int { - 1dx} }}\]

\[I.F. = {e^{ - x}}\]

Therefore, solution of the given differential equation is given by the relation,

\[x\left( {I.F.} \right) = \int {\left( {Q \times I.F.} \right)dy}  + C\]

\[y{e^{ - x}} = \int {\left( {x - 5} \right){e^{ - x}}dx}  + C\]

\[y{e^{ - x}} = \left( {x - 5} \right)\int {{e^{ - x}}dx}  - \int {\left( {\dfrac{d}{{dx}}\left( {x - 5} \right)\int {{e^{ - x}}dx} } \right)} dx + C\]

\[y{e^{ - x}} = \left( {5 - x} \right){e^{ - x}} + \int {{e^{ - x}}dx}  + C\]

\[y{e^{ - x}} = \left( {5 - x} \right){e^{ - x}} - {e^{ - x}} + C\]

\[y{e^{ - x}} = \left( {4 - x} \right){e^{ - x}} + C\]

\[y = 4 - x + C{e^x}\]

\[y + x - 4 = C{e^x}......\left( 1 \right)\]

As equation is passing through \[\left( {0,2} \right)\] .

Therefore,

\[0 + 2 - 4 = C{e^0}\]

\[ - 2 = C\]

Substituting the value of \[ - 2 = C\] in equation 1.

\[y + x - 4 =  - 2{e^x}\]

This the required equation of curve passing through \[\left( {0,2} \right)\] .

18. The Integrating Factor of the differential equation \[x\dfrac{{dy}}{{dx}} - y = 2{x^2}\] is 

\[\left( A \right){e^{ - x}}\]

\[\left( B \right){e^{ - y}}\]

\[\left( C \right)\dfrac{1}{x}\]

\[\left( D \right)x\]

Ans: On rearranging the given equation can be written as,

\[\dfrac{{dy}}{{dx}} - \dfrac{y}{x} = 2x\]

The given equation is in the form of \[\dfrac{{dy}}{{dx}} + Py = Q\]

Where, \[P =  - \dfrac{1}{x}\] and \[Q = 2x\]

Calculating integration factor,

\[I.F. = {e^{\int {Pdx} }}\]

\[I.F. = {e^{\int { - \dfrac{1}{x}dx} }}\]

\[I.F. = {e^{ - \log x}}\]

\[I.F. = {e^{\log \dfrac{1}{x}}}\]

\[I.F. = \dfrac{1}{x}\]

Therefore, the correct option is \[\left( C \right)\] .

18. The Integrating Factor of the differential equation \[\left( {1 - {y^2}} \right)\dfrac{{dx}}{{dy}} + yx = ay\left( { - 1 < y < 1} \right)\] is

\[\left( A \right)\dfrac{1}{{{y^2} - 1}}\]

\[\left( B \right)\dfrac{1}{{\sqrt {{y^2} - 1} }}\]

\[\left( C \right)\dfrac{1}{{1 - {y^2}}}\]

\[\left( D \right)\dfrac{1}{{\sqrt {1 - {y^2}} }}\]

Ans: On rearranging the given equation can be written as,

\[\dfrac{{dx}}{{dy}} + \dfrac{{xy}}{{1 - {y^2}}} = \dfrac{{ay}}{{1 - {y^2}}}\]

The given equation is in the form of \[\dfrac{{dx}}{{dy}} + Px = Q\]

Where, \[P = \dfrac{y}{{1 - {y^2}}}\] and \[Q = \dfrac{{ay}}{{1 - {y^2}}}\]

Calculating integration factor,

\[I.F. = {e^{\int {Pdy} }}\]

\[I.F. = {e^{\int {\dfrac{y}{{1 - {y^2}}}dy} }}\]

\[\]\[I.F. = {e^{\log \left( {\dfrac{1}{{\sqrt {1 - {y^2}} }}} \right)}}\]

\[I.F. = \dfrac{1}{{\sqrt {1 - {y^2}} }}\]

Therefore, the correct option is \[\left( D \right)\]

Conclusion

NCERT Solutions for Class 12 Maths Chapter 9 - Differential Equations Exercise 9.5 offers a thorough approach to solving linear differential equations. The detailed, step-by-step solutions provided help clarify complex concepts and methods, such as using integrating factors. These solutions are designed to enhance understanding and problem-solving skills, making them a valuable resource for effective exam preparation. Download the FREE PDF to strengthen your grasp of linear differential equations and boost your confidence for upcoming exams.

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