CBSE 2024 · Region 1 · Set 2 · Q27 · 3 marks
Find the general solution of the differential equation $\displaystyle \frac{d \mathrm{y}}{d \mathrm{x}}=\frac{\mathrm{x}^{2}+\mathrm{y}^{2}}{2 \mathrm{x} \mathrm{y}}$.
Marking-scheme solution
$$\begin{aligned}
\text { Put } \mathrm{y} & =\mathrm{vx} \text { so that } \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{v}+\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}} \\
& \mathrm{v}+\mathrm{x} \frac{d \mathrm{v}}{d \mathrm{x}}=\frac{\mathrm{x}^{2}+\mathrm{v}^{2} \mathrm{x}^{2}}{2 \mathrm{x} \mathrm{v} \mathrm{x}} \\
& \int \frac{1}{\mathrm{x}} d \mathrm{x}=\int \frac{2 \mathrm{v}}{1-\mathrm{v}^{2}} d \mathrm{v} \\
& \Rightarrow \log |\mathrm{x}|=-\log \left|1-\mathrm{v}^{2}\right|+\log \mathrm{C} \\
& \log \left|\mathrm{x}\left(1-\mathrm{v}^{2}\right)\right|=\log \mathrm{C} \\
& \Rightarrow \mathrm{x}\left(1-\frac{\mathrm{y}^{2}}{\mathrm{x}^{2}}\right)=\mathrm{C} \text { or } \mathrm{x}^{2}-\mathrm{y}^{2}=\mathrm{C} \mathrm{x}
\end{aligned}
$$
Differential EquationsMethods of Solving First Order, First Degree Differential EquationsApplyshort_answermedium
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CBSE Class 12 Mathematics past-paper question from the 2024board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.