NCERT Solutions for Class 12 Maths Chapter 9: Differential Equations - Exercise 9.5

EXERCISE 9.5

Refer for exercise 9.5 in the PDF

1. Solve the differential equation \[\left( {{x^2} + xy} \right)dy = \left( {{x^2} + {y^2}} \right)dy\]

Ans: After rearranging the given equation we get,

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

It is a homogeneous differential equation.

To solve this problem, we will make the substitution.

\[y = vx\]

Differentiating equation w.r.t. \[x\], we get

\[\dfrac{{dy}}{{dx}} = v + x\dfrac{{dv}}{{dx}}\]

Substituting \[y = vx\] and \[\dfrac{{dy}}{{dx}}\] in the above equation, we get

\[v + x\dfrac{{dv}}{{dx}} = \dfrac{{{x^2} + {{\left( {vx} \right)}^2}}}{{{x^2} + x\left( {vx} \right)}}\]

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

\[v + x\dfrac{{dv}}{{dx}} = \dfrac{{1 + {v^2}}}{{1 + v}}\]

\[x\dfrac{{dv}}{{dx}} = \dfrac{{1 + {v^2}}}{{1 + v}} - v\]

\[x\dfrac{{dv}}{{dx}} = \dfrac{{1 + {v^2} - v - {v^2}}}{{1 + v}}\]

\[x\dfrac{{dv}}{{dx}} = \dfrac{{1 - v}}{{1 + v}}\]

\[dfrac{1}{x}dx = \dfrac{{1 + v}}{{1 - v}}dv\]

Taking integration on both side,

\[\int {\dfrac{1}{x}dx}  = \int {\dfrac{{1 + v}}{{1 - v}}dv} \]

\[\log x + C = \int {\dfrac{{1 + v + 1 - 1}}{{1 - v}}dv} \]

\[\log x + C = \int {\dfrac{{2 + 1 - v}}{{1 - v}}dv} \]

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

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

Substituting the value of \[v = \dfrac{y}{x}\].

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

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

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

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

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

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

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

This is the required differential equation.

Where \[C = {e^{ - C}}\].

2. Solve the differential equation \[y' = \dfrac{{x + y}}{x}\]

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

On rearranging the equation, we get

\[\dfrac{{dy}}{{dx}} = 1 + \dfrac{y}{x}\]

It is a homogeneous differential equation.

To solve this problem, we will make the substitution.

\[y = vx\]

Differentiating equation w.r.t. \[x\], we get

\[\dfrac{{dy}}{{dx}} = v + x\dfrac{{dv}}{{dx}}\]

Substituting \[y = vx\] and \[\dfrac{{dy}}{{dx}}\] in the above equation, we get

\[v + x\dfrac{{dv}}{{dx}} = 1 + v\]

\[dv = \dfrac{1}{x}dx\]

Taking integration on both side,

\[\int {1.dv}  = \int {\dfrac{1}{x}dx} \]

\[v = \log x + C\]

Substituting the value of \[v = \dfrac{y}{x}\].

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

\[y = x\log x + Cx\]

This is the required differential equation.

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

Ans: After rearranging the given equation we get,

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

It is a homogeneous differential equation.

To solve this problem, we will make the substitution.

\[y = vx\]

Differentiating equation w.r.t. \[x\], we get

\[\dfrac{{dy}}{{dx}} = v + x\dfrac{{dv}}{{dx}}\]

Substituting \[y = vx\] and \[\dfrac{{dy}}{{dx}}\] in the above equation, we get

\[v + x\dfrac{{dv}}{{dx}} = \dfrac{{x + vx}}{{x - vx}}\]

\[v + x\dfrac{{dv}}{{dx}} = \dfrac{{1 + v}}{{1 - v}}\]

\[x\dfrac{{dv}}{{dx}} = \dfrac{{1 + v}}{{1 - v}} - v\]

\[x\dfrac{{dv}}{{dx}} = \dfrac{{1 + v - v + {v^2}}}{{1 - v}}\]

\[x\dfrac{{dv}}{{dx}} = \dfrac{{1 + {v^2}}}{{1 - v}}\]

\[\dfrac{{1 - v}}{{1 + {v^2}}}dv = \dfrac{1}{x}dx\]

Taking integration on both side,

\[\int {\dfrac{{1 - v}}{{1 + {v^2}}}dv}  = \int {\dfrac{1}{x}dx} \]

\[\int {\dfrac{1}{{1 + {v^2}}}} dv - \int {\dfrac{v}{{1 + {v^2}}}} dv = \int {\dfrac{1}{x}dx} ......\left( 1 \right)\]

Let \[I = \int {\dfrac{v}{{1 + {v^2}}}} dv\]

Now, let \[1 + {v^2} = t\]

Differentiating equation w.r.t. \[v\], we get

\[  2vdv = dt \] 

 \[ vdv = \dfrac{{dt}}{2} \]

Substituting \[vdv = \dfrac{{dt}}{2}\] in the above equation, we get

\[I = \int {\dfrac{1}{2t}} dt\]

Substituting this value in equation \[\left( 1 \right)\] .

Therefore,

\[\int {\dfrac{1}{{1 + {v^2}}}} dv - \int {\dfrac{1}{2t}} dt = \int {\dfrac{1}{x}dx} \]

\[{\tan ^{ - 1}}v - \dfrac{1}{2}\log t = \log x + C\]

Substituting the value of \[1 + {v^2} = t\] and \[v = \dfrac{y}{x}\] .

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

After rearranging the given equation we get,

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

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

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

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

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

This is the required differential equation.

4. Solve the differential equation \[\left( {{x^2} - {y^2}} \right)dx + 2xydy = 0\]

Ans: After rearranging the given equation we get,

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

It is a homogeneous differential equation.

To solve this problem, we will make the substitution.

\[y = vx\]

Differentiating equation w.r.t. \[x\], we get

\[\dfrac{{dy}}{{dx}} = v + x\dfrac{{dv}}{{dx}}\]

Substituting \[y = vx\] and \[\dfrac{{dy}}{{dx}}\] in the above equation, we get

\[v + x\dfrac{{dv}}{{dx}} =  - \dfrac{{{x^2} - {{\left( {vx} \right)}^2}}}{{2x.vx}}\]

\[v + x\dfrac{{dv}}{{dx}} =  - \dfrac{{1 - {v^2}}}{{2v}}\]

\[x\dfrac{{dv}}{{dx}} =  - \dfrac{{1 - {v^2}}}{{2v}} - v\]

\[x\dfrac{{dv}}{{dx}} = \dfrac{{ - 1 + {v^2} - 2{v^2}}}{{2v}}\]

\[x\dfrac{{dv}}{{dx}} = \dfrac{{ - 1 - {v^2}}}{{2v}}\]

\[ - \dfrac{{2v}}{{1 + {v^2}}}dv = \dfrac{1}{x}dx\]

\[\dfrac{{2v}}{{1 + {v^2}}}dv =  - \dfrac{1}{x}dx\]

Taking integration on both side,

\[\int {\dfrac{{2v}}{{1 + {v^2}}}dv}  =  - \int {\dfrac{1}{x}dx} ......\left( 1 \right)\]

Now, let \[1 + {v^2} = t\]

Differentiating equation w.r.t. \[v\], we get

\[2vdv = dt\]

Substituting \[2vdv = dt\] in equation \[\left( 1 \right)\], we get

\[\int {\dfrac{1}{t}dt}  =  - \int {\dfrac{1}{x}dx} \]

\[\log t =  - \log x + C\]

Substituting the value of \[1 + {v^2} = t\] and \[v = \dfrac{y}{x}\].

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

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

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

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

\[\dfrac{{{x^2} + {y^2}}}{x} = K\]

\[{x^2} + {y^2} = Kx\]

This is the required differential equation.

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

Ans: After rearranging the given equation we get,

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

It is a homogeneous differential equation.

To solve this problem, we will make the substitution.

\[y = vx\]

Differentiating equation w.r.t. \[x\], we get

\[\dfrac{{dy}}{{dx}} = v + x\dfrac{{dv}}{{dx}}\]

Substituting \[\dfrac{{dy}}{{dx}}\] in the above equation, we get

\[v + x\dfrac{{dv}}{{dx}} = 1 - 2{v^2} + v\]

\[x\dfrac{{dv}}{{dx}} = 1 - 2{v^2}\]

\[\dfrac{1}{{1 - 2{v^2}}}dv = \dfrac{1}{x}dx\]

Taking integration on both side,

\[\int {\dfrac{1}{{1 - 2{v^2}}}dv}  = \int {\dfrac{1}{x}dx} \]

\[\int {\dfrac{1}{{{1^2} - {{\left( {\sqrt 2 v} \right)}^2}}}dv}  = \int {\dfrac{1}{x}dx} \]

On integrating using standard trigonometric identity we get,

\[\dfrac{1}{{\sqrt 2 }}.\dfrac{1}{{1.2}}.\log \left| {\dfrac{{1 + \sqrt 2 v}}{{1 - \sqrt 2 v}}} \right| = \log \left| x \right| + C\]

Substituting the value of \[v = \dfrac{y}{x}\].

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

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

This is the required differential equation.

6. Solve the differential equation \[xdy - ydx = \sqrt {{x^2} + {y^2}} dx\]

Ans: After rearranging the given equation we get,

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

\[\dfrac{{dy}}{{dx}} = \sqrt {1 + \dfrac{{{y^2}}}{{{x^2}}}}  + \dfrac{y}{x}\]

It is a homogeneous differential equation.

To solve this problem, we will make the substitution.

\[y = vx\]

Differentiating equation w.r.t. \[x\], we get

\[\dfrac{{dy}}{{dx}} = v + x\dfrac{{dv}}{{dx}}\]

Substituting \[y = vx\] and \[\dfrac{{dy}}{{dx}}\] in the above equation, we get

\[v + x\dfrac{{dv}}{{dx}} = \sqrt {1 + {v^2}}  + v\]

\[x\dfrac{{dv}}{{dx}} = \sqrt {1 + {v^2}} \]

\[\dfrac{1}{{\sqrt {1 + {v^2}} }}dv = \dfrac{1}{x}dx\]

Taking integration on both side,

\[\int {\dfrac{1}{{\sqrt {1 + {v^2}} }}dv}  = \int {\dfrac{1}{x}dx} \]

Using \[\int {\dfrac{1}{{\sqrt {{x^2} + {a^2}} }} = \log \left| {x + \sqrt {{x^2} + {a^2}} } \right|}  + C\], we get

\[\log \left| {v + \sqrt {1 + {v^2}} } \right| = \log x + \log C\]

Substituting the value of \[v = \dfrac{y}{x}\].

\[\log \left| {\dfrac{y}{x} + \sqrt {1 + {{\left( {\dfrac{y}{x}} \right)}^2}} } \right| = \log xC\]

\[\dfrac{y}{x} + \sqrt {1 + {{\left( {\dfrac{y}{x}} \right)}^2}}  = xC\]

\[\dfrac{y}{x} + \sqrt {{{\dfrac{{{x^2} + y}}{{{x^2}}}}^2}}  = xC\]

\[\dfrac{y}{x} + \dfrac{{\sqrt {{x^2} + {y^2}} }}{x} = xC\]

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

This is the required differential equation.

7. Solve the differential equation \[\left\{ {x\cos \left( {\dfrac{y}{x}} \right) + y\sin \left( {\dfrac{y}{x}} \right)} \right\}ydx = \left\{ {y\sin \left( {\dfrac{y}{x}} \right) - x\cos \left( {\dfrac{y}{x}} \right)} \right\}xdy\]

Ans: After rearranging the given equation we get,

\[\dfrac{{dy}}{{dx}} = \dfrac{{\left\{ {x\cos \left( {\dfrac{y}{x}} \right) + y\sin \left( {\dfrac{y}{x}} \right)} \right\}y}}{{\left\{ {y\sin \left( {\dfrac{y}{x}} \right) - x\cos \left( {\dfrac{y}{x}} \right)} \right\}x}}\]

It is a homogeneous differential equation.

To solve this problem, we will make the substitution.

\[y = vx\]

Differentiating equation w.r.t. \[x\], we get

\[\dfrac{{dy}}{{dx}} = v + x\dfrac{{dv}}{{dx}}\]

Substituting \[y = vx\] and \[\dfrac{{dy}}{{dx}}\] in the above equation, we get

\[v + x\dfrac{{dv}}{{dx}} = \dfrac{{\left\{ {x\cos \left( v \right) + vx\sin \left( v \right)} \right\}vx}}{{\left\{ {vx\sin \left( v \right) - x\cos \left( v \right)} \right\}x}}\]

\[x\dfrac{{dv}}{{dx}} = \dfrac{{\left\{ {\cos \left( v \right) + v\sin \left( v \right)} \right\}v}}{{\left\{ {v\sin \left( v \right) - \cos \left( v \right)} \right\}}} - v\]

\[x\dfrac{{dv}}{{dx}} = \dfrac{{v\cos \left( v \right) + {v^2}\sin \left( v \right) - {v^2}\sin \left( v \right) + v\cos \left( v \right)}}{{v\sin \left( v \right) - \cos \left( v \right)}}\]

\[x\dfrac{{dv}}{{dx}} = \dfrac{{2v\cos \left( v \right)}}{{v\sin \left( v \right) - \cos \left( v \right)}}\]

\[\dfrac{{v\sin \left( v \right) - \cos \left( v \right)}}{{2v\cos \left( v \right)}}dv = \dfrac{1}{x}dx\]

Taking integration on both side,

\[\int {\dfrac{{v\sin \left( v \right) - \cos \left( v \right)}}{{2v\cos \left( v \right)}}dv}  = \int {\dfrac{1}{x}dx} \]

\[\int {\dfrac{{v\sin \left( v \right)}}{{2v\cos \left( v \right)}}dv - \int {\dfrac{{\cos \left( v \right)}}{{2v\cos \left( v \right)}}dv} }  = \int {\dfrac{1}{x}dx} \]

\[\dfrac{1}{2}\int {\tan vdv - \dfrac{1}{2}\int {\dfrac{1}{v}dv} }  = \int {\dfrac{1}{x}dx} \]

\[\dfrac{1}{2}\log \sec v - \dfrac{1}{2}\log v = \log x + \log C\]

\[\dfrac{1}{2}\left( {\log \sec v - \log v} \right) = \log xC\]

\[\log \dfrac{{\sec v}}{v} = 2\log xC\]

\[\log \dfrac{{\sec v}}{v} = \log {\left( {xC} \right)^2}\]

\[\dfrac{{\sec v}}{v} = {\left( {xC} \right)^2}\]

Substituting the value of \[v = \dfrac{y}{x}\].

\[\dfrac{{\sec \left( {\dfrac{y}{x}} \right)}}{{\dfrac{y}{x}}} = {\left( {xC} \right)^2}\]

\[\dfrac{x}{y}\sec \left( {\dfrac{y}{x}} \right) = {\left( {xC} \right)^2}\]

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

\[\dfrac{1}{{yx}}.\dfrac{1}{{\cos \left( {\dfrac{y}{x}} \right)}} = {C^2}\]

\[\dfrac{1}{{{C^2}}} = yx\cos \left( {\dfrac{y}{x}} \right)\]

\[yx\cos \left( {\dfrac{y}{x}} \right) = K\]

This is the required differential equation.

8. Solve the differential equation \[x\dfrac{{dy}}{{dx}} - y + x\sin \left( {\dfrac{y}{x}} \right) = 0\]

Ans: After rearranging the given equation we get,

\[\dfrac{{dy}}{{dx}} = \dfrac{{y - x\sin \left( {\dfrac{y}{x}} \right)}}{x}\]

\[\dfrac{{dy}}{{dx}} = \dfrac{y}{x} - \sin \left( {\dfrac{y}{x}} \right)\]

It is a homogeneous differential equation.

To solve this problem, we will make the substitution.

\[y = vx\]

Differentiating equation w.r.t. \[x\], we get

\[\dfrac{{dy}}{{dx}} = v + x\dfrac{{dv}}{{dx}}\]

Substituting \[y = vx\] and \[\dfrac{{dy}}{{dx}}\] in the above equation, we get

\[v + x\dfrac{{dv}}{{dx}} = \dfrac{{vx}}{x} - \sin \left( {\dfrac{{vx}}{x}} \right)\]

\[v + x\dfrac{{dv}}{{dx}} = v - \sin v\]

\[x\dfrac{{dv}}{{dx}} =  - \sin v\]

\[\dfrac{1}{{\sin v}}dv =  - \dfrac{1}{x}dx\]

Taking integration on both side,

\[\int {\dfrac{1}{{\sin v}}dv}  =  - \int {\dfrac{1}{x}dx} \]

\[\int {\cos ecvdv}  =  - \int {\dfrac{1}{x}dx} \]

\[\log \left( {\cos ecv - \cot v} \right) =  - \log x + \log C\]

\[\log \left( {\cos ecv - \cot v} \right) = \log \dfrac{C}{x}\]

\[\cos ecv - \cot v = \dfrac{C}{x}\]

Substituting the value of \[v = \dfrac{y}{x}\].

\[\cos ec\left( {\dfrac{y}{x}} \right) - \cot \left( {\dfrac{y}{x}} \right) = \dfrac{C}{x}\]

\[\dfrac{1}{{\sin \left( {\dfrac{y}{x}} \right)}} - \dfrac{{\cos \left( {\dfrac{y}{x}} \right)}}{{\sin \left( {\dfrac{y}{x}} \right)}} = \dfrac{C}{x}\]

\[1 - \cos \left( {\dfrac{y}{x}} \right) = \dfrac{C}{x}\sin \left( {\dfrac{y}{x}} \right)\]

\[x\left( {1 - \cos \left( {\dfrac{y}{x}} \right)} \right) = C\sin \left( {\dfrac{y}{x}} \right)\]

This is the required differential equation.

9. Solve the differential equation \[ydx + x\log \left( {\dfrac{y}{x}} \right)dy - 2xdy = 0\]

Ans: After rearranging the given equation we get,

\[\dfrac{{dy}}{{dx}} = \dfrac{{ - y}}{{x\log \left( {\dfrac{y}{x}} \right) - 2x}}\]

\[\dfrac{{dy}}{{dx}} = \dfrac{y}{{2x - x\log \left( {\dfrac{y}{x}} \right)}}\]

It is a homogeneous differential equation.

To solve this problem, we will make the substitution.

\[y = vx\]

Differentiating equation w.r.t. \[x\], we get

\[\dfrac{{dy}}{{dx}} = v + x\dfrac{{dv}}{{dx}}\]

Substituting \[y = vx\] and \[\dfrac{{dy}}{{dx}}\] in the above equation, we get

\[v + x\dfrac{{dv}}{{dx}} = \dfrac{{vx}}{{2x - x\log \left( {\dfrac{{vx}}{x}} \right)}}\]

\[v + x\dfrac{{dv}}{{dx}} = \dfrac{v}{{2 - \log v}}\]

\[x\dfrac{{dv}}{{dx}} = \dfrac{v}{{2 - \log v}} - v\]

\[x\dfrac{{dv}}{{dx}} = \dfrac{{v - 2v + v\log v}}{{2 - \log v}}\]

\[x\dfrac{{dv}}{{dx}} = \dfrac{{ - v + v\log v}}{{2 - \log v}}\]

\[\dfrac{{2 - \log v}}{{ - v + v\log v}}dv = \dfrac{1}{x}dx\]

\[\dfrac{{2 - \log v}}{{v\left( {\log v - 1} \right)}}dv = \dfrac{1}{x}dx\]

\[\dfrac{{1 - \left( {\log v - 1} \right)}}{{v\left( {\log v - 1} \right)}}dv = \dfrac{1}{x}dx\]

\[\dfrac{1}{{v\left( {\log v - 1} \right)}}dv - \dfrac{{\left( {\log v - 1} \right)}}{{v\left( {\log v - 1} \right)}}dv = \dfrac{1}{x}dx\]

\[\dfrac{1}{{v\left( {\log v - 1} \right)}}dv - \dfrac{1}{v}dv = \dfrac{1}{x}dx\]

Taking integration on both side,

\[\int {\dfrac{1}{{v\left( {\log v - 1} \right)}}dv}  - \int {\dfrac{1}{v}dv}  = \int {\dfrac{1}{x}dx} \]                                            ………. (1)

Let \[I = \int {\dfrac{1}{{v\left( {\log v - 1} \right)}}dv} \]

Put \[\log v - 1 = t\]

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

\[\dfrac{1}{v} = \dfrac{{dt}}{{dv}}\]

\[\dfrac{1}{v}dv = dt\]

Put this value in above equation and we get

\[I = \int {\dfrac{1}{t}dt} \]

Put this in equation \[\left( 1 \right)\]

\[\int {\dfrac{1}{t}dt}  - \int {\dfrac{1}{v}dv}  = \int {\dfrac{1}{x}dx} \]

\[\log t - \log v = \log x + \log c\]

Substituting the value of \[\log v - 1 = t\] and \[v = \dfrac{y}{x}\] .

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

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

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

\[\log \left( {\dfrac{y}{x}} \right) - 1 = \dfrac{y}{x}.xC\]

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

This is the required differential equation.

10. Solve the differential equation \[\left( {1 + {e^{\dfrac{x}{y}}}} \right)dx + {e^{\dfrac{x}{y}}}\left( {1 - \dfrac{x}{y}} \right)dy = 0\]

Ans: After rearranging the given equation we get,

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

It is a homogeneous differential equation.

To solve this problem, we will make the substitution.

\[x = vy\]

Differentiating equation w.r.t. \[y\], we get

\[\dfrac{{dx}}{{dy}} = v + y\dfrac{{dv}}{{dy}}\]

Substituting \[x = vy\] and \[\dfrac{{dx}}{{dy}}\] in the above equation, we get

\[v + y\dfrac{{dv}}{{dy}} = \dfrac{{ - {e^{\dfrac{{vy}}{y}}}\left( {1 - \dfrac{{vy}}{y}} \right)}}{{1 + {e^{\dfrac{{vy}}{y}}}}}\]

\[v + y\dfrac{{dv}}{{dy}} = \dfrac{{ - {e^v}\left( {1 - v} \right)}}{{1 + {e^v}}}\]

\[v + y\dfrac{{dv}}{{dy}} = \dfrac{{ - {e^v} + v{e^v}}}{{1 + {e^v}}}\]

\[y\dfrac{{dv}}{{dy}} = \dfrac{{ - {e^v} + v{e^v}}}{{1 + {e^v}}} - v\]

\[y\dfrac{{dv}}{{dy}} = \dfrac{{ - {e^v} + v{e^v} - v - v{e^v}}}{{1 + {e^v}}}\]

\[y\dfrac{{dv}}{{dy}} = \dfrac{{ - {e^v} - v}}{{1 + {e^v}}}\]

\[y\dfrac{{dv}}{{dy}} =  - \left[ {\dfrac{{{e^v} + v}}{{1 + {e^v}}}} \right]\]

\[\left[ {\dfrac{{1 + {e^v}}}{{{e^v} + v}}} \right]dv =  - \dfrac{1}{y}dy\]

Taking integration on both side,

\[\int {\left[ {\dfrac{{1 + {e^v}}}{{{e^v} + v}}} \right]dv}  =  - \int {\dfrac{1}{y}dy} \]

On integrating both side,

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

Substituting the value of \[v = \dfrac{x}{y}\].

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

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

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

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

This is the required differential equation.

11. Solve the differential equation \[\left( {x + y} \right)dy + \left( {x - y} \right)dx = 0;{\text{ }}y = 1;x = 1\]

Ans: After rearranging the given equation we get,

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

It is a homogeneous differential equation.

To solve this problem, we will make the substitution.

\[y = vx\]

Differentiating equation w.r.t. \[x\], we get

\[\dfrac{{dy}}{{dx}} = v + x\dfrac{{dv}}{{dx}}\]

Substituting \[y = vx\] and \[\dfrac{{dy}}{{dx}}\] in the above equation, we get

\[v + x\dfrac{{dv}}{{dx}} =  - \dfrac{{\left( {x - vx} \right)}}{{\left( {x + vx} \right)}}\]

\[v + x\dfrac{{dv}}{{dx}} =  - \dfrac{{\left( {1 - v} \right)}}{{\left( {1 + v} \right)}}\]

\[x\dfrac{{dv}}{{dx}} =  - \dfrac{{\left( {1 - v} \right)}}{{\left( {1 + v} \right)}} - v\]

\[x\dfrac{{dv}}{{dx}} = \dfrac{{ - 1 + v - v - {v^2}}}{{1 + v}}\]

\[x\dfrac{{dv}}{{dx}} = \dfrac{{ - 1 - {v^2}}}{{1 + v}}\]

\[\dfrac{{1 + v}}{{1 + {v^2}}}dv =  - \dfrac{1}{x}dx\]

Taking integration on both side,

\[\int {\dfrac{{1 + v}}{{1 + {v^2}}}dv}  =  - \int {\dfrac{1}{x}dx} \]

\[\int {\dfrac{1}{{1 + {v^2}}}dv}  + \int {\dfrac{v}{{1 + {v^2}}}dv}  =  - \int {\dfrac{1}{x}dx} \]

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

Substituting the value of \[v = \dfrac{y}{x}\].

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

\[y = 1\]

When

\[x = 1\]

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

\[\dfrac{\pi }{4} + \dfrac{1}{2}\log \left( 2 \right) = C\]

Therefore, the final solution becomes,

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

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

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

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

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

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

This is the required differential equation.

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

Ans: After rearranging the given equation we get,

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

It is a homogeneous differential equation.

To solve this problem, we will make the substitution.

\[y = vx\]

Differentiating equation w.r.t. \[x\], we get

\[\dfrac{{dy}}{{dx}} = v + x\dfrac{{dv}}{{dx}}\]

Substituting \[y = vx\] and \[\dfrac{{dy}}{{dx}}\] in the above equation, we get

\[v + x\dfrac{{dv}}{{dx}} =  - \dfrac{{xvx + {{\left( {xv} \right)}^2}}}{{{x^2}}}\]

\[v + x\dfrac{{dv}}{{dx}} =  - \dfrac{{v{x^2} + {x^2}{v^2}}}{{{x^2}}}\]

\[v + x\dfrac{{dv}}{{dx}} =  - v - {v^2}\]

\[x\dfrac{{dv}}{{dx}} =  - v - {v^2} - v\]

\[x\dfrac{{dv}}{{dx}} =  - 2v - {v^2}\]

\[\dfrac{1}{{2v + {v^2}}}dv =  - \dfrac{1}{x}dx\]

Taking integration on both side,

\[\int {\dfrac{1}{{2v + {v^2}}}dv}  =  - \int {\dfrac{1}{x}dx} \]

\[\int {\dfrac{1}{{v\left( {2 + v} \right)}}dv}  =  - \int {\dfrac{1}{x}dx} \]

Dividing and multiplying above equation by 2.

\[\dfrac{1}{2}\int {\dfrac{2}{{v\left( {2 + v} \right)}}dv}  =  - \int {\dfrac{1}{x}dx} \]

\[\dfrac{1}{2}\int {\dfrac{{2 + v - v}}{{v\left( {2 + v} \right)}}dv}  =  - \int {\dfrac{1}{x}dx} \]

\[\dfrac{1}{2}\int {\dfrac{{2 + v}}{{v\left( {2 + v} \right)}}dv}  - \dfrac{1}{2}\int {\dfrac{v}{{v\left( {2 + v} \right)}}dv}  =  - \int {\dfrac{1}{x}dx} \]

\[\dfrac{1}{2}\int {\dfrac{1}{v}dv}  - \dfrac{1}{2}\int {\dfrac{1}{{\left( {2 + v} \right)}}dv}  =  - \int {\dfrac{1}{x}dx} \]

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

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

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

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

\[\dfrac{v}{{2 + v}} = {\left( {\dfrac{C}{x}} \right)^2}\]

Substituting the value of \[v = \dfrac{y}{x}\].

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

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

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

\[y = 1\]

When

\[x = 1\]

\[\dfrac{1}{{2.1 + 1}} = {\left( {\dfrac{C}{1}} \right)^2}\]

\[\dfrac{1}{3} = {C^2}\]

Therefore, the final solution becomes,

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

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

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

\[\dfrac{{y{x^2}}}{{2x + y}} = \dfrac{1}{3}\]

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

This is the required differential equation.

13. Solve the differential equation \[\left[ {x{{\sin }^2}\left( {\dfrac{y}{x}} \right) - y} \right]dx + xdy = 0\]; \[y = \dfrac{\pi }{4}\] when \[x = 1\].

Ans: After rearranging the given equation we get,

\[\dfrac{{dy}}{{dx}} =  - \dfrac{{\left[ {x{{\sin }^2}\left( {\dfrac{y}{x}} \right) - y} \right]}}{x}\]

It is a homogeneous differential equation.

To solve this problem, we will make the substitution.

\[y = vx\]

Differentiating equation w.r.t. \[x\], we get

\[\dfrac{{dy}}{{dx}} = v + x\dfrac{{dv}}{{dx}}\]

Substituting \[y = vx\] and \[\dfrac{{dy}}{{dx}}\] in the above equation, we get

\[v + x\dfrac{{dv}}{{dx}} =  - \dfrac{{\left[ {x{{\sin }^2}\left( {\dfrac{{vx}}{x}} \right) - vx} \right]}}{x}\]

\[v + x\dfrac{{dv}}{{dx}} =  - {\sin ^2}\left( v \right) + v\]

\[x\dfrac{{dv}}{{dx}} =  - {\sin ^2}\left( v \right) + v - v\]

\[x\dfrac{{dv}}{{dx}} =  - {\sin ^2}\left( v \right)\]

\[\dfrac{1}{{{{\sin }^2}\left( v \right)}}dv =  - \dfrac{1}{x}dx\]

\[\cos e{c^2}vdv =  - \dfrac{1}{x}dx\]

Taking integration on both side,

\[\int {\cos e{c^2}vdv}  =  - \int {\dfrac{1}{x}dx} \]

\[ - \cot v =  - \log x - \log C\]

\[\cot v = \log x + \log C\]

\[\cot v = \log xC\]

Substituting the value of \[v = \dfrac{y}{x}\].

\[\cot \left( {\dfrac{y}{x}} \right) = \log xC\]

\[y = \dfrac{\pi }{4}\]

When

\[x = 1\]

\[\cot \left( {\dfrac{{\dfrac{\pi }{4}}}{1}} \right) = \log 1C\]

\[\cot \left( {\dfrac{\pi }{4}} \right) = \log C\]

\[1 = \log C\]

\[{e^1} = C\]

\[e = C\]

Therefore, final solution becomes,

\[\cot \left( {\dfrac{y}{x}} \right) = \log \left| {xe} \right|\]

This is the required differential equation.

14. Solve the differential equation \[\dfrac{{dy}}{{dx}} - \dfrac{y}{x} + \cos ec\left( {\dfrac{y}{x}} \right) = 0\]; \[y = 0\] when \[x = 1\].

Ans: After rearranging the given equation we get,

\[\dfrac{{dy}}{{dx}} = \dfrac{y}{x} - \cos ec\left( {\dfrac{y}{x}} \right)\]

It is a homogeneous differential equation.

To solve this problem, we will make the substitution.

\[y = vx\]

Differentiating equation w.r.t. \[x\], we get

\[\dfrac{{dy}}{{dx}} = v + x\dfrac{{dv}}{{dx}}\]

Substituting \[y = vx\] and \[\dfrac{{dy}}{{dx}}\] in the above equation, we get

\[v + x\dfrac{{dv}}{{dx}} = \dfrac{{vx}}{x} - \cos ec\left( {\dfrac{{vx}}{x}} \right)\]

\[v + x\dfrac{{dv}}{{dx}} = v - \cos ec\left( v \right)\]

\[x\dfrac{{dv}}{{dx}} =  - \cos ec\left( v \right)\]

\[\dfrac{1}{{\cos ec\left( v \right)}}dv =  - \dfrac{1}{x}dx\]

\[\sin vdv =  - \dfrac{1}{x}dx\]

Taking integration on both side,

\[\int {\sin vdv}  =  - \int {\dfrac{1}{x}dx} \]

\[ - \cos v =  - \log x + C\]

Substituting the value of \[v = \dfrac{y}{x}\].

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

\[y = 0\]

When

\[x = 1\]

\[ - \cos \left( {\dfrac{0}{1}} \right) =  - \log 1 + C\]

\[ - 1 = C\]

Therefore, the final solution becomes,

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

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

\[\cos \left( {\dfrac{y}{x}} \right) = \log x + \log e\]

\[\cos \left( {\dfrac{y}{x}} \right) = \log \left| {xe} \right|\]

This is the required differential equation.

15. Solve the differential equation \[2xy + {y^2} - 2{x^2}\dfrac{{dy}}{{dx}} = 0\]; \[y = 2\] when \[x = 1\].

Ans: After rearranging the given equation we get,

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

It is a homogeneous differential equation.

To solve this problem, we will make the substitution.

\[y = vx\]

Differentiating equation w.r.t. \[x\], we get

\[\dfrac{{dy}}{{dx}} = v + x\dfrac{{dv}}{{dx}}\]

Substituting \[y = vx\] and \[\dfrac{{dy}}{{dx}}\] in the above equation, we get

\[v + x\dfrac{{dv}}{{dx}} = \dfrac{{2xvx + {{\left( {vx} \right)}^2}}}{{2{x^2}}}\]

\[v + x\dfrac{{dv}}{{dx}} = \dfrac{{2{x^2}v + {v^2}{x^2}}}{{2{x^2}}}\]

\[v + x\dfrac{{dv}}{{dx}} = \dfrac{{2v + {v^2}}}{2}\]

\[x\dfrac{{dv}}{{dx}} = \dfrac{{2v + {v^2}}}{2} - v\]

\[x\dfrac{{dv}}{{dx}} = \dfrac{{2v + {v^2} - 2v}}{2}\]

\[x\dfrac{{dv}}{{dx}} = \dfrac{{{v^2}}}{2}\]

\[\dfrac{2}{{{v^2}}}dv = \dfrac{1}{x}dx\]

Taking integration on both side,

\[\int {\dfrac{2}{{{v^2}}}dv}  = \int {\dfrac{1}{x}dx} \]

\[ - \dfrac{2}{v} = \log x + C\]

Substituting the value of \[v = \dfrac{y}{x}\].

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

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

\[y = 2\]

When

\[x = 1\]

\[ - \dfrac{{2.1}}{2} = \log 1 + C\]

\[ - 1 = C\]

Therefore, final solution becomes,

\[ - \dfrac{{2x}}{y} = \log x - 1\]

\[\dfrac{{2x}}{y} = 1 - \log x\]

\[y = \dfrac{{2x}}{{1 - \log x}}:x \ne e\] 

This is the required differential equation.

16. A homogeneous differential equation of the from \[\dfrac{{dx}}{{dy}} = h\left( {\dfrac{x}{y}} \right)\]can be solved by

making the substitution.

\[ \left( A \right)y = vx{\text{                      }}\left( B \right)v = yx\]

  \[\left( C \right)x = vy{\text{                      }}\left( D \right)x = v\]

Ans: As \[h\left( {\dfrac{x}{y}} \right)\] is function of \[\dfrac{x}{y}\]

Therefore, we have to substitute, \[x = vy\] .

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

17. Which of the following is a homogeneous differential equation?

\[\left( A \right)\left( {4x + 6y + 5} \right)dy - \left( {3y + 2x + 4} \right)dx = 0\]

\[\left( B \right)\left( {xy} \right)dx - \left( {{x^3} + {y^3}} \right)dy = 0\]

\[\left( C \right)\left( {{x^3} + 2{y^2}} \right)dx + 2xydy = 0\]

\[\left( D \right){y^2}dx + \left( {{x^2} - xy - {y^2}} \right)dy = 0\]

Ans: The correct option is \[\left( D \right)\] .

Explanation:

\[{y^2}dx + \left( {{x^2} - xy - {y^2}} \right)dy = 0\]

After rearranging the given equation we get,

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

Let \[f\left( {x,y} \right) =  - \dfrac{{{x^2} - xy - {y^2}}}{{{y^2}}}\]

Now, put \[x = kx\] and \[y = ky\] 

\[f\left( {kx,ky} \right) =  - \dfrac{{{{\left( {kx} \right)}^2} - kxky - {{\left( {ky} \right)}^2}}}{{{{\left( {ky} \right)}^2}}}\]

\[f\left( {kx,ky} \right) =  - \dfrac{{{k^2}}}{{{k^2}}}\dfrac{{{x^2} - xy - {y^2}}}{{{y^2}}}\]

\[f\left( {kx,ky} \right) = {k^0}\left( { - \dfrac{{{x^2} - xy - {y^2}}}{{{y^2}}}} \right)\]

\[f\left( {kx,ky} \right) = {k^0}f\left( {x,y} \right)\]

Hence, the given differential equation is homogeneous.

NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Exercise 9.5

Opting for the NCERT solutions for Ex 9.5 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 9.5 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 9.5 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 9.5, 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. 

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NCERT Solutions for Class 12 Maths PDF Download

• Chapter 1 - Relations and Functions

• Chapter 2 - Inverse Trigonometric Functions

• Chapter 3 - Matrices

• Chapter 4 - Determinants

• Chapter 5 - Continuity and Differentiability

• Chapter 6 - Application of Derivatives

• Chapter 7 - Integrals

• Chapter 8 - Application of Integrals

• Chapter 9 - Differential Equations

• Chapter 10 - Vector Algebra

• Chapter 11 - Three Dimensional Geometry

• Chapter 12 - Linear Programming

• Chapter 13 - Probability

Students can also download the following additional study material -

• Chapter 9: Important Questions

• Chapter 9: Revision Notes

• Chapter 9: Formulas

• Chapter 9: NCERT Exemplar Solutions

• Chapter 9: RD Sharma Solutions