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

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Download Free PDF of Differential Equations Exercise 9.4 Solutions for Class 12 Maths

Differential Equations form the heart of Class 12 Maths, and mastering Exercise 9.4 is essential for your CBSE board journey. Here, you will strengthen your understanding of variable separable methods and the logic behind first order, first degree equations—both directly mapped in the latest Class 12 Maths syllabus.

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With a chapter weightage of 7 marks, questions from this topic appear every year. If you ever find yourself searching for “exercise 9.4 class 12”, you are aiming for reliable, stepwise NCERT solutions that remove confusion from both the method and integration steps. Here, integrating factor steps and solved differential equation examples are explained in the exact way you would apply them in board exams.


Every solution is checked for full board compliance and detailed reasoning, curated by Vedantu’s academic experts. Use this as your smart revision aid to build confidence—quickly and clearly—before your final exam.

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Access NCERT Solutions for Maths Class 12 Chapter 9 - Differential Equations

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.


Class 12 Maths Chapter 9: Exercises Breakdown

S.No.

Chapter 9 - Differential Equations Exercises in PDF Format

1

Class 12 Maths Chapter 9 Exercise 9.1 - 12 Questions & Solutions (10 Short Answers, 2 MCQs)

2

Class 12 Maths Chapter 9 Exercise 9.2 - 12 Questions & Solutions (10 Short Answers, 2 MCQs)

3

Class 12 Maths Chapter 9 Exercise 9.3 - 12 Questions & Solutions (5 Short Answers, 5 Long Answers, 2 MCQs)

4

Class 12 Maths Chapter 9 Exercise 9.5 - 17 Questions & Solutions (15 Short Answers, 2 MCQs)



CBSE Class 12 Maths Chapter 9 Other Study Materials



NCERT Solutions for Class 12 Maths | Chapter-wise List

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FAQs on CBSE Class 12 Maths Chapter 9 Differential Equations – NCERT Solutions 2025-26

1. What is the method used to solve Exercise 9.4 in Class 12 Maths?

Exercise 9.4 in Class 12 Maths is solved mainly using the variable separable method for first order, first degree differential equations.

Key steps:

  • Identify the equation’s type (first order, first degree).
  • Separate the variables (put all y-terms and dy on one side; x-terms and dx on the other).
  • Integrate both sides with respect to their variables.
  • Add the constant of integration C.
  • Express the solution in general or specified form.

This approach follows the NCERT/CBSE pattern and covers exam-focused methods such as integration practice and separating variables.

2. How can I download Exercise 9.4 NCERT Solutions as a PDF?

You can easily download Class 12 Maths Chapter 9 Exercise 9.4 NCERT Solutions PDF for offline study and revision.

Steps to Download:

  • Go to the page titled NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.4.
  • Look for the Download PDF or highlighted button section.
  • Click to download; registration or login is not required on most trusted educational sites.

Having a PDF lets you revise and practise variable separable differential equations quickly before exams, even without internet.

3. Are Vedantu's Class 12 Maths solutions Board-marking scheme compliant?

Yes, Vedantu's Class 12 Maths NCERT solutions are fully aligned with the CBSE Board marking scheme.

How the solutions match Board guidelines:

  • Every step is shown as per CBSE’s stepwise marking.
  • The latest 2025 syllabus is followed strictly.
  • Key formulas, definitions, and integration techniques are clearly highlighted.
  • Accurate and concise answers increase scoring chances.

This ensures you get reliable, exam-ready answers for Differential Equations and all exercises in Chapter 9.

4. What are the tips to avoid common calculation mistakes in Differential Equations?

To avoid calculation mistakes in Differential Equations (especially in Exercise 9.4):

Follow these tips:

  • Always separate variables carefully; double-check rearrangement.
  • Use the correct integration rules and mark boundaries clearly.
  • Do not forget to add the integration constant "C" at the end.
  • Write each step; skip none, as per CBSE marking.
  • After final answer, quickly verify by differentiating to check correctness.

Practice with stepwise solutions to strengthen accuracy and minimize exam errors.

5. How is Exercise 9.4 important for CBSE Board and entrance exams?

Exercise 9.4 is crucial for both CBSE Board and entrance exams like JEE and NEET.

Importance includes:

  • Covers the frequently tested topic: first order, first degree differential equations.
  • Makes up to 7 marks in CBSE Maths Boards (2025 syllabus).
  • NEET and JEE often have direct or application-based questions from this exercise.
  • Helps solidify concepts needed for higher calculus in college entrance.

Mastering this exercise boosts confidence and increases scores in board and competitive exams.

6. Where can I find stepwise explanations and reasoning for each Ex 9.4 problem?

Stepwise, Board-oriented explanations for every Exercise 9.4 question are available in expert-verified NCERT solutions.

Features of these solutions:

  • Each step—variable separation, integration, and simplification—is explained.
  • Hints, formulas, and short explanations boost understanding.
  • Solutions are shown exactly in the format expected for CBSE checking.
  • You may also find revision notes and key points on trusted learning sites.

This approach strengthens concept clarity and prepares you for full marks in exam.

7. Is class 12 maths very tough?

Class 12 Maths is considered moderate if you focus on understanding core concepts, practising stepwise NCERT solutions, and revising regularly.

Tips to manage difficulty:

  • Study chapter-wise solutions, especially for high-weightage chapters like Differential Equations.
  • Practice by solving past years’ questions for each exercise.
  • Download and use PDF revision materials before exams.
With regular effort and clear guidance, most students find Class 12 Maths manageable.

8. Which is the hardest chapter in class 12 maths?

The difficulty varies, but chapters like Probability, Application of Integrals, and at times Differential Equations pose challenges due to multistep solutions and conceptual depth.

Common tough chapters:

  • Probability (Chapter 13)
  • Application of Integrals (Chapter 8)
  • Differential Equations (Chapter 9) – for integration steps
  • Three-dimensional Geometry (Chapter 11)
Focused stepwise solutions and summary notes can help make even the hardest chapters easier to master.

9. What is the general form of a differential equation Class 12?

The general form of a first order, first degree differential equation in Class 12 is:

dy/dx = f(x)g(y) (Variable separable form)

Or, for linear equations:
dy/dx + P(x)y = Q(x)

These forms allow you to use variable separable method or integrating factor method as per Exercise 9.4 and the NCERT syllabus.

10. What are the key points to remember while solving Class 12 Maths Chapter 9 Exercise 9.4?

To score full marks in Exercise 9.4, keep these key points in mind:

  • Write the equation in separated variable form: dy/dx = f(x)g(y).
  • Move all y terms (and dy) to one side; all x terms (and dx) to the other.
  • Integrate both sides properly; watch for constants.
  • Don’t forget to add the constant of integration C.
  • Box your final answer and check clarity as per the CBSE marking scheme.
This ensures your answers meet exam-oriented and syllabus-based requirements for maximum marks.