CBSE 2026 · Region 2 · Set 1 · Q29 · 3 marks
Solve the following differential equation : $\displaystyle x \frac{\mathrm{dy}}{\mathrm{d} x}=\mathrm{y}-x \sin ^{2}\left(\frac{\mathrm{y}}{x}\right)$, given that $\displaystyle \mathrm{y}(1)=\frac{\pi}{6}$Find the general solution of the differential equation : $\displaystyle \mathrm{y} \log \mathrm{y} \frac{\mathrm{d} x}{\mathrm{dy}}+x=\frac{2}{\mathrm{y}}$.
Solve the following differential equation : $\displaystyle x \frac{\mathrm{dy}}{\mathrm{d} x}=\mathrm{y}-x \sin ^{2}\left(\frac{\mathrm{y}}{x}\right)$, given that $\displaystyle \mathrm{y}(1)=\frac{\pi}{6}$
Find the general solution of the differential equation : $\displaystyle \mathrm{y} \log \mathrm{y} \frac{\mathrm{d} x}{\mathrm{dy}}+x=\frac{2}{\mathrm{y}}$.
Marking-scheme solution
Here, $\displaystyle \dfrac{dy}{dx}=\dfrac{\mathrm{y}}{x}-\sin^{2}\left(\dfrac{\mathrm{y}}{x}\right)$
Put $\displaystyle \dfrac{\mathrm{y}}{x}=v \Rightarrow \dfrac{dy}{dx}=v+x \dfrac{dv}{dx}$
Differential equation reduces to
$\displaystyle v+x \dfrac{dv}{dx}=v-\sin^{2} v$
i.e. $\displaystyle x \dfrac{dv}{dx}=-\sin^{2} v$
$\displaystyle \Rightarrow-\int \operatorname{cosec}^{2} v\, dv=\int \dfrac{dx}{x}$
$\displaystyle \Rightarrow \cot v=\log |x|+C$
$\displaystyle \Rightarrow \cot\left(\dfrac{\mathrm{y}}{x}\right)=\log |x|+C$
Now $\displaystyle \mathrm{y}(1)=\dfrac{\pi}{6}$ gives $\displaystyle C=\sqrt{3}$
Required solution is : $\displaystyle \cot\left(\dfrac{\mathrm{y}}{x}\right)=\log |x|+\sqrt{3}$
Given differential equation can be written as
$\displaystyle \dfrac{dx}{dy}+\dfrac{1}{\mathrm{y} \log \mathrm{y}} \cdot x=\dfrac{2}{\mathrm{y}^{2} \log \mathrm{y}}$
I.F. $\displaystyle =\mathrm{e}^{\int \frac{1}{\mathrm{y} \log \mathrm{y}} dy}=\mathrm{e}^{\log (\log \mathrm{y})}=\log \mathrm{y}$
Solution is given by
$\displaystyle x \times \log \mathrm{y}=\int \dfrac{2}{\mathrm{y}^{2} \log \mathrm{y}} \times \log \mathrm{y}\, dy=\int \dfrac{2}{\mathrm{y}^{2}}\, dy$
$\displaystyle \Rightarrow x \log \mathrm{y}=-\dfrac{2}{\mathrm{y}}+C$ or $\displaystyle x \mathrm{y} \log \mathrm{y}=-2+C \mathrm{y}$
Differential EquationsMethods of Solving First Order, First Degree Differential EquationsApplyshort_answerhard
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CBSE Class 12 Mathematics past-paper question from the 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.