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Mathematics · 2026 · 3 marks
Differential EquationsMethods of Solving First Order, First Degree Differential Equations3 marksApplyshort answer
CBSE 2026 · Region 3 · Set 2 · Q27
(a) Find the general solution of the differential equation ( x^2+ y^2) d y = x y d x.
(a)
Find the general solution of the differential equation \[\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right) d \mathrm{y}=\mathrm{x} \mathrm{y} d \mathrm{x} . \]
OR
(b)
Find the particular solution of the differential equation $\displaystyle \frac{\mathrm{dy}}{\mathrm{dx}}-3 \mathrm{y} \cot \mathrm{x}=\sin 2 \mathrm{x}$, given that $\displaystyle \mathrm{y}=2$ when $\displaystyle \mathrm{x}=\frac{\pi}{2}$.
Marking-scheme solution
$\displaystyle \dfrac{dy}{dx}=\dfrac{xy}{x^{2}+y^{2}}$
Put $\displaystyle y=vx$ and $\displaystyle \dfrac{dy}{dx}=v+x \dfrac{dv}{dx}$
Hence, $\displaystyle v+x \dfrac{dv}{dx}=\dfrac{v}{1+v^{2}} \Rightarrow x \dfrac{dv}{dx}=-\dfrac{v^{3}}{1+v^{2}}$
$\displaystyle \Rightarrow \int \dfrac{v^{2}+1}{v^{3}} dv=-\int \dfrac{dx}{x}$
$\displaystyle \Rightarrow \log |v|-\dfrac{1}{2 v^{2}}=-\log |x|+c$
$\displaystyle \Rightarrow \log\left|\dfrac{y}{x}\right|-\dfrac{x^{2}}{2 y^{2}}=-\log |x|+c$ or $\displaystyle \log y=\dfrac{x^{2}}{2 y^{2}}+c$
$\displaystyle \dfrac{dy}{dx}-3 y \cot x=\sin 2x$
I.F. $\displaystyle =e^{-3 \int \cot x\, dx}=\dfrac{1}{\sin^{3} x}$ or $\displaystyle \operatorname{cosec}^{3} x$
General solution is given by $\displaystyle \dfrac{y}{\sin^{3} x}=\int \dfrac{\sin 2x}{\sin^{3} x} dx$
$\displaystyle \Rightarrow y \operatorname{cosec}^{3} x=-2 \operatorname{cosec} x+c$
Put $\displaystyle x=\dfrac{\pi}{2}$ and $\displaystyle y=2$ ; we get $\displaystyle c=4$
Hence, particular solution is given by
$\displaystyle y \operatorname{cosec}^{3} x=-2 \operatorname{cosec} x+4$ or $\displaystyle y=-2 \sin^{2} x+4 \sin^{3} x$
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CBSE Class 12 Mathematics past-paper question from the 2026 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.