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Mathematics · 2023 · 3 marks
Differential EquationsMethods of Solving First Order, First Degree Differential Equations3 marksApplyshort answer
CBSE 2023 · Region 1 · Set 2 · Q26
(a) Find the general solution of the differential equation: (d x)/(d y)=(e^x / y (x/y-1))/(1+e^x / y).
(a)
Find the general solution of the differential equation : \[\frac{d \mathrm{x}}{d \mathrm{y}}=\frac{e^{\mathrm{x} / \mathrm{y}}\left(\dfrac{\mathrm{x}}{\mathrm{y}}-1\right)}{1+e^{\mathrm{x} / \mathrm{y}}} . \]
OR
(b)
Find the particular solution of the differential equation $\displaystyle \frac{\mathrm{dy}}{\mathrm{dx}}+\cot \mathrm{x} \cdot \mathrm{y}=\cos ^{2} \mathrm{x}$, given that when $\displaystyle \mathrm{x}=\frac{\pi}{2}, \mathrm{y}=0$.
Marking-scheme solution
Let $\displaystyle x = vy \Rightarrow \dfrac{dx}{dy} = v + y\dfrac{dv}{dy}$
Substituting in the given differential equation, we get
\[v + y\frac{dv}{dy} = \frac{e^{v}(v-1)}{e^{v}+1} \Rightarrow y\frac{dv}{dy} = -\frac{(e^{v}+v)}{e^{v}+1} \]
\[\Rightarrow \frac{e^{v}+1}{e^{v}+v}\,dv = -\frac{dy}{y} \]
Integrating we get
$\displaystyle \log\left|e^{v}+v\right| = -\log|y| + \log C$
\[\Rightarrow e^{x/y} + \frac{x}{y} = \frac{C}{y} \quad \text{or} \quad y\,e^{x/y} + x = C \]
Integrating factor is $\displaystyle e^{\int \cot x\,dx} = e^{\log \sin x} = \sin x$
Solution is $\displaystyle y \sin x = \displaystyle\int \cos^{2} x \, \sin x \, dx + C$
\[\Rightarrow y \sin x = -\frac{\cos^{3} x}{3} + C \]
$\displaystyle x = \dfrac{\pi}{2},\ y = 0 \Rightarrow C = 0$
$\displaystyle \therefore$ particular solution is $\displaystyle y \sin x = -\dfrac{\cos^{3} x}{3}$ or $\displaystyle y = -\dfrac{\cos^{3} x}{3}\cdot \operatorname{cosec} x$
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CBSE Class 12 Mathematics past-paper question from the 2023 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.