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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 3 · Q30
(a) Find the general solution of the differential equation ( x^2- y^2) d x +2 x y d y =0
(a)
Find the general solution of the differential equation \[\left(\mathrm{x}^{2}-\mathrm{y}^{2}\right) d \mathrm{x}+2 \mathrm{x} \mathrm{y} d \mathrm{y}=0 \]
OR
(b)
Solve the differential equation \[\sin \mathrm{x} \cos \mathrm{y} d \mathrm{x}+\cos \mathrm{x} \sin \mathrm{y} d \mathrm{y}=0 \text {, given that } \mathrm{y}=\frac{\pi}{4} \text { when } \mathrm{x}=0 \text {. } \]
Marking-scheme solution
$\displaystyle \dfrac{dy}{dx}=\dfrac{y^{2}-x^{2}}{2 x y}$
Put $\displaystyle y=v x$ and $\displaystyle \dfrac{dy}{dx}=v+x \dfrac{dv}{dx}$
Thus, $\displaystyle v+x \dfrac{dv}{dx}=\dfrac{v^{2}-1}{2 v}$
$\displaystyle \Rightarrow x \dfrac{dv}{dx}=-\dfrac{1+v^{2}}{2 v}$
$\displaystyle \Rightarrow \int \dfrac{2 v}{1+v^{2}} dv=-\int \dfrac{dx}{x}$
$\displaystyle \Rightarrow \log\left|1+v^{2}\right|=-\log |x|+\log |c|$
$\displaystyle \Rightarrow 1+v^{2}=\dfrac{c}{x}$
$\displaystyle \Rightarrow x^{2}+y^{2}=c x$
$\displaystyle \sin x \cos y\, dx+\cos x \sin y\, dy=0$
$\displaystyle \Rightarrow \dfrac{dy}{dx}=-\dfrac{\sin x \cos y}{\cos x \sin y}$
$\displaystyle \Rightarrow \int \tan y\, dy=-\int \tan x\, dx$
$\displaystyle \Rightarrow \log |\sec y|=-\log |\sec x|+\log |c|$
$\displaystyle \Rightarrow \sec x \sec y=c$
Put $\displaystyle x=0, y=\dfrac{\pi}{4}$ we get $\displaystyle c=\sqrt{2}$
Hence particular solution is given by $\displaystyle \sec x \sec y=\sqrt{2}$
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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.