CBSE 2026 · Region 1 · Set 3 · Q30 · 3 marks
Find the general solution of the differential equation $\displaystyle \left(\mathrm{y}^{2}-x^{2}\right) \mathrm{d} x=2 x \mathrm{y} \mathrm{dy}$Find the particular solution of the differential equation $\displaystyle \left(1+\mathrm{e}^{2 x}\right) \mathrm{dy}+\left(1+\mathrm{y}^{2}\right) \mathrm{e}^{x} \mathrm{~d} x=0$, given that $\displaystyle \mathrm{y}(1)=0$.
Find the general solution of the differential equation $\displaystyle \left(\mathrm{y}^{2}-x^{2}\right) \mathrm{d} x=2 x \mathrm{y} \mathrm{dy}$
Find the particular solution of the differential equation $\displaystyle \left(1+\mathrm{e}^{2 x}\right) \mathrm{dy}+\left(1+\mathrm{y}^{2}\right) \mathrm{e}^{x} \mathrm{~d} x=0$, given that $\displaystyle \mathrm{y}(1)=0$.
Marking-scheme solution
Given differential equation is $\displaystyle \dfrac{dy}{dx} = \dfrac{1}{2}\left[\dfrac{\mathrm{y}}{x} - \dfrac{x}{\mathrm{y}}\right]$
Put $\displaystyle \mathrm{y} = vx \Rightarrow \dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$
The given differential equation becomes $\displaystyle v + x\dfrac{dv}{dx} = \dfrac{1}{2}\left[v - \dfrac{1}{v}\right]$
$\displaystyle \Rightarrow \dfrac{2v\,dv}{1+v^2} = -\dfrac{dx}{x}$
$\displaystyle \Rightarrow \int \dfrac{2v\,dv}{1+v^2} = -\int \dfrac{dx}{x}$
$\displaystyle \Rightarrow \log(1+v^2) = -\log|x| + \log C$
$\displaystyle \Rightarrow \log(1+v^2)|x| = \log C$
$\displaystyle \Rightarrow (1+v^2)x = C$
$\displaystyle \Rightarrow x^2 + \mathrm{y}^2 = Cx$
Given differential equation is $\displaystyle \dfrac{dy}{1+\mathrm{y}^2} =- \dfrac{\mathrm{e}^x}{1+\mathrm{e}^{2x}}\,dx$
$\displaystyle \Rightarrow \int \dfrac{dy}{1+\mathrm{y}^2} = -\int \dfrac{\mathrm{e}^x}{1+\mathrm{e}^{2x}}\,dx$
$\displaystyle \Rightarrow \tan^{-1}\mathrm{y} = -\int \dfrac{1}{1+t^2}\,dt, \quad \left[\mathrm{e}^x = t \Rightarrow \mathrm{e}^x dx = dt\right]$
$\displaystyle \Rightarrow \tan^{-1}\mathrm{y} + \tan^{-1}(\mathrm{e}^x) = C$
When $\displaystyle x = 1, \mathrm{y} = 0 \Rightarrow C = \tan^{-1}(\mathrm{e})$
Hence, the required particular solution is $\displaystyle \tan^{-1}\mathrm{y} + \tan^{-1}(\mathrm{e}^x) = \tan^{-1}(\mathrm{e})$
Differential EquationsMethods of Solving First Order, First Degree Differential EquationsApplyshort_answermedium
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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.