NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.4

EXERCISE 9.4

Refer for exercise 9.4 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.

Conclusion

In conclusion, Exercise 9.4 of Chapter 9, "Differential Equations," is a critical part of your understanding of differential equations, focusing on advanced methods of solving these equations. This class 12 ex 9.4 deals with higher-order differential equations and various techniques such as the method of undetermined coefficients and variation of parameters. Through consistent practice, you have strengthened your ability to approach complex differential equations methodically, applying appropriate solution strategies to each unique problem.

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