CBSE 2023 · Region 3 · Set 1 · Q26 · 3 marks
Find the general solution of the differential equation : \[\frac{d}{d \mathrm{x}}\left(\mathrm{x} \mathrm{y}^{2}\right)=2 \mathrm{y}\left(1+\mathrm{x}^{2}\right) \]Solve the following differential equation : \[\mathrm{x} \mathrm{e}^{\frac{\mathrm{y}}{\mathrm{x}}}-\mathrm{y}+\mathrm{x} \frac{d \mathrm{y}}{d \mathrm{x}}=0 \]
Find the general solution of the differential equation : \[\frac{d}{d \mathrm{x}}\left(\mathrm{x} \mathrm{y}^{2}\right)=2 \mathrm{y}\left(1+\mathrm{x}^{2}\right) \]
Solve the following differential equation : \[\mathrm{x} \mathrm{e}^{\frac{\mathrm{y}}{\mathrm{x}}}-\mathrm{y}+\mathrm{x} \frac{d \mathrm{y}}{d \mathrm{x}}=0 \]
Marking-scheme solution
(a)
Given differential equation is
$$\begin{aligned}
& $\displaystyle 2$ \mathrm{xy} \frac{d \mathrm{y}}{d \mathrm{x}}+\mathrm{y}^{$\displaystyle 2$}=$\displaystyle 2$ \mathrm{y}\left($\displaystyle 1$+\mathrm{x}^{$\displaystyle 2$}\right)
& \Rightarrow \frac{d \mathrm{y}}{d \mathrm{x}}+\frac{\mathrm{y}}{$\displaystyle 2$ \mathrm{x}}=\frac{1}{\mathrm{x}}+\mathrm{x}
\end{aligned}
$$Integrating factor $\displaystyle =\mathrm{e}^{\int \frac{1}{2 \mathrm{x}} d \mathrm{x}}=\mathrm{e}^{\log \sqrt{\mathrm{x}}}=\sqrt{\mathrm{x}}$
Solution is given by $\displaystyle \mathrm{y} \sqrt{\mathrm{x}}=\int\left(\frac{1}{\sqrt{\mathrm{x}}}+\mathrm{x}^{\frac{3}{2}}\right) \mathrm{dx}$
$$\Rightarrow \mathrm{y} \sqrt{\mathrm{x}}=$\displaystyle 2$ \sqrt{\mathrm{x}}+\frac{$\displaystyle 2$ \mathrm{x}^{\frac{5}{2}}}{$\displaystyle 5$}+\mathrm{C} \text {, or } \mathrm{y}=$\displaystyle 2$+\frac{$\displaystyle 2$ \mathrm{x}^{$\displaystyle 2$}}{$\displaystyle 5$}+\frac{\mathrm{C}}{\sqrt{\mathrm{x}}}
$$(b) Given differential equation is $\displaystyle \frac{d \mathrm{y}}{d \mathrm{x}}=\frac{\mathrm{y}}{\mathrm{x}}-\mathrm{e}^{\frac{\mathrm{y}}{\mathrm{x}}}$
Let $\displaystyle \mathrm{y}=\mathrm{vx} \Rightarrow \frac{d \mathrm{y}}{d \mathrm{x}}=\mathrm{v}+\mathrm{x} \frac{d \mathrm{v}}{d \mathrm{x}}$
The given equation becomes $\displaystyle \mathrm{v}+\mathrm{x} \frac{d \mathrm{v}}{d \mathrm{x}}=\mathrm{v}-\mathrm{e}^{\mathrm{v}}$
$$\Rightarrow-\mathrm{e}^{-\mathrm{v}} \mathrm{dv}=\frac{d \mathrm{x}}{\mathrm{x}}
$$Integrating both sides, we get
$$\begin{aligned}
& \mathrm{e}^{-\mathrm{V}}=\log |\mathrm{x}|+\mathrm{C}
& \Rightarrow \mathrm{e}^{-\frac{\mathrm{y}}{\mathrm{x}}}=\log |\mathrm{x}|+\mathrm{C}
\end{aligned}
Differential EquationsMethods of Solving First Order, First Degree Differential EquationsApplyshort_answermedium
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CBSE Class 12 Mathematics past-paper question from the 2023board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.