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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 1 · Q29
(a) Find the general solution of the differential equation 2 x^2 dy/dx= y^2+2 xy.
(a)
Find the general solution of the differential equation $\displaystyle 2 \mathrm{x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{y}^{2}+2 \mathrm{xy}$.
OR
(b)
Find a particular solution of the differential equation $\displaystyle (\mathrm{x}+1) \frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{e}^{-\mathrm{y}}-1$, given that $\displaystyle \mathrm{y}=0$ when $\displaystyle \mathrm{x}=0$.
Marking-scheme solution
$\displaystyle 2 x^{2} \dfrac{dy}{dx}=y^{2}+2 x y \Rightarrow \dfrac{dy}{dx}=\dfrac{1}{2}\left(\dfrac{y}{x}\right)^{2}+\dfrac{y}{x}$
Put $\displaystyle y=v x$ and $\displaystyle \dfrac{dy}{dx}=v+x \dfrac{dv}{dx}$
Hence, $\displaystyle v+x \dfrac{dv}{dx}=\dfrac{1}{2} v^{2}+v$
$\displaystyle \Rightarrow \int \dfrac{1}{v^{2}} dv=\int \dfrac{1}{2 x} dx$
$\displaystyle \Rightarrow-\dfrac{1}{v}=\dfrac{1}{2} \log x+c$
$\displaystyle \Rightarrow-\dfrac{x}{y}=\dfrac{1}{2} \log x+c$
$\displaystyle (x+1) \dfrac{dy}{dx}=2 e^{-y}-1 \Rightarrow \int \dfrac{e^{y}}{2-e^{y}} dy=\int \dfrac{1}{x+1} dx$
Put $\displaystyle 2-e^{y}=t \Rightarrow e^{y} dy=-dt$
Hence, $\displaystyle -\int \dfrac{dt}{t}=\log |x+1|$
$\displaystyle \Rightarrow-\log |t|=\log |x+1|-\log |c|$
$\displaystyle \Rightarrow(x+1)\left(2-e^{y}\right)=c$
Put $\displaystyle x=y=0 ; \quad c=1$
Hence, particular solution is given by
$\displaystyle (x+1)\left(2-e^{y}\right)=1$ or $\displaystyle \log |x+1|+\log \left|2-e^{-y}\right|=0$
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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.