NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations (Ex 9.6) Exercise 9.6

NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations (Ex 9.6) Exercise 9.6

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Access NCERT Solutions for Class 12 Chapter 9 – Differential Equations part-1

Access NCERT Solutions for Class 12 Chapter 9 – Differential Equations

Exercise 9.6

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 \[\left( D \right)\]


NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Exercise 9.6

Opting for the NCERT solutions for Ex 9.6 Class 12 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 9.6 Class 12 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 12 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 12 Maths Chapter 9 Exercise 9.6 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 12 Maths Chapter 9 Exercise 9.6, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it. 

Do not delay any more. Download the NCERT solutions for Class 12 Maths Chapter 9 Exercise 9.6 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well. 

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