CBSE 2026 · Region 1 · Set 1 · Q29 · 3 marks
Find the general solution of the following differential equation : \[x^{2} \frac{\mathrm{dy}}{\mathrm{dx}}=x^{2}+x \mathrm{y}+\mathrm{y}^{2}
\]Find the particular solution of the differential equation$\displaystyle x \mathrm{y} \frac{\mathrm{dy}}{\mathrm{d} x}=(x+2)(\mathrm{y}+2)$, given that $\displaystyle \mathrm{y}(1)=-1$.
Find the general solution of the following differential equation : \[x^{2} \frac{\mathrm{dy}}{\mathrm{dx}}=x^{2}+x \mathrm{y}+\mathrm{y}^{2}
\]
Find the particular solution of the differential equation
$\displaystyle x \mathrm{y} \frac{\mathrm{dy}}{\mathrm{d} x}=(x+2)(\mathrm{y}+2)$, given that $\displaystyle \mathrm{y}(1)=-1$.
Official answer
From CBSE’s own marking scheme for this paper.
(a)
y = x·tan(ln|x| + C) or equivalent form; (b) ln|y+$\displaystyle 2$| - ln|x| = 2ln|y+$\displaystyle 1$| + ln($\displaystyle 3$) or equivalent implicit form
Marking-scheme solution
Given differential equation is :$\displaystyle \dfrac{dy}{dx} = 1+\dfrac{\mathrm{y}}{x}+\left(\dfrac{\mathrm{y}}{x}\right)^2$
Put $\displaystyle \mathrm{y}=vx \Rightarrow \dfrac{dy}{dx} = v+x\dfrac{dv}{dx}$
The given differential equation becomes $\displaystyle x\dfrac{dv}{dx} = 1+v^2$
$\displaystyle \Rightarrow \dfrac{dv}{1+v^2} = \dfrac{dx}{x}$
$\displaystyle \Rightarrow \int\dfrac{dv}{1+v^2} = \int\dfrac{dx}{x}$
$\displaystyle \Rightarrow \tan^{-1}v = \log|x|+C$
General solution: $\displaystyle \tan^{-1}\left(\dfrac{\mathrm{y}}{x}\right) = \log|x|+C$
Given differential equation is: $\displaystyle \mathrm{xy} \frac{\mathrm{dy}}{\mathrm{d} x}=(x+2)(\mathrm{y}+2)$
$\displaystyle \Rightarrow \dfrac{\mathrm{y}}{\mathrm{y}+2}\,dy = \dfrac{x+2}{x}\,dx$
$\displaystyle \Rightarrow \int\left(1-\dfrac{2}{\mathrm{y}+2}\right)dy = \int\left(1+\dfrac{2}{x}\right)dx$
$\displaystyle \Rightarrow \mathrm{y}-2\log|\mathrm{y}+2| = x+2\log|x|+C$
At $\displaystyle x=1,\mathrm{y}=-1$: $\displaystyle C=-2$
Particular solution: $\displaystyle \mathrm{y}-2\log|\mathrm{y}+2| = x+2\log|x|-2$, or $\displaystyle \mathrm{y} = x+2\log|x(\mathrm{y}+2)|-2$
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.