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Mathematics · 2026 · 5 marks
Differential EquationsMethods of Solving First Order, First Degree Differential Equations5 marksApplylong answer
CBSE 2026 · Region 4 · Set 1 · Q34
(a) Solve the differential equation y e^y dx =( y^3+2 x e^y ) dy, when y (0)=1.
(a)
Solve the differential equation y $\displaystyle \mathrm{e}^{\mathrm{y}} \mathrm{dx}=\left(\mathrm{y}^{3}+2 \mathrm{x} \mathrm{e}^{\mathrm{y}}\right) \mathrm{dy}$, when $\displaystyle \mathrm{y}(0)=1$.
OR
(b)
Find the general solution of the differential equation \[\left(\mathrm{x}^{3}-3 \mathrm{x} \mathrm{y}^{2}\right) d \mathrm{x}=\left(\mathrm{y}^{3}-3 \mathrm{x}^{2} \mathrm{y}\right) d \mathrm{y} . \]
Marking-scheme solution
Given differential eq. can be written as $\displaystyle \dfrac{dx}{dy}-\dfrac{2}{y} x=y^{2} e^{-y}$
Integrating factor $\displaystyle =e^{\int \frac{-2}{y} d y}=\dfrac{1}{y^{2}}$
General solution: $\displaystyle x\left(\dfrac{1}{y^{2}}\right)=\int \dfrac{1}{y^{2}} \cdot y^{2} e^{-y} d y$
$\displaystyle x\left(\dfrac{1}{y^{2}}\right)=-e^{-y}+c$ or $\displaystyle x=y^{2}\left(-e^{-y}+c\right)$
Now, when $\displaystyle x=0, y=1: 0=-e^{-1}+c \Rightarrow c=e^{-1}$
Required particular solution is, $\displaystyle x=y^{2}\left(e^{-1}-e^{-y}\right)$
$\displaystyle \dfrac{dy}{dx}=\dfrac{x^{3}-3 x y^{2}}{y^{3}-3 x^{2} y}$
Put $\displaystyle y=v x \Rightarrow \dfrac{dy}{dx}=v+x \dfrac{dv}{dx}$, so
$\displaystyle v+x \dfrac{dv}{dx}=\dfrac{1-3 v^{2}}{v^{3}-3 v} \Rightarrow x \dfrac{dv}{dx}=\dfrac{1-3 v^{2}}{v^{3}-3 v}-v=\dfrac{1-v^{4}}{v^{3}-3 v}$
$\displaystyle \Rightarrow \int \dfrac{3 v-v^{3}}{v^{4}-1} d v=\int \dfrac{1}{x} d x$
$\displaystyle \Rightarrow \dfrac{3}{4} \log\left|\dfrac{v^{2}-1}{v^{2}+1}\right|-\dfrac{1}{4} \log\left|v^{4}-1\right|=\log x+c$
$\displaystyle \therefore \dfrac{3}{4} \log\left|\dfrac{y^{2}-x^{2}}{y^{2}+x^{2}}\right|-\dfrac{1}{4} \log\left|\dfrac{y^{4}-x^{4}}{x^{4}}\right|=\log x+c$
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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.