Q.3 (B) (i) Solve the following differential equation (x²-y²) dx+2xy dy=0

Answer: x²+y²=cx dy/dx=y²-x²/2xy . This is a homogeneous equation. Put y=vx , so dy/dx=v+xdv/dx . v+xdv/dx=v²x²-x²/2vx²=v²-1/2v xdv/dx=v²-1/2v-v=-(1+v²)/2v Separating the variables: 2v/1+v² dv=-dx/x Integrating: log(1+v²)=-log x+log c⇒ x(1+v²)=c Put v=y/x : x(1+y²/x²)=c⇒ x²+y²=cx , which is the general solution.

Solve the following differential equation\((x^2-y^2)\,dx+2xy\,dy=0\)

Q.3 (B) (i) Solve the following differential equation

\((x^2-y^2)\,dx+2xy\,dy=0\)

Doubtly Asked 28 Oct 2026 Board exam · March 2026 hsc 12th Save WhatsApp Report

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Answer: \(x^2+y^2=cx\)

\(\frac{dy}{dx}=\frac{y^2-x^2}{2xy}\). This is a homogeneous equation. Put \(y=vx\), so \(\frac{dy}{dx}=v+x\frac{dv}{dx}\).

\(v+x\frac{dv}{dx}=\frac{v^2x^2-x^2}{2vx^2}=\frac{v^2-1}{2v}\)

\(x\frac{dv}{dx}=\frac{v^2-1}{2v}-v=\frac{-(1+v^2)}{2v}\)

Separating the variables: \(\frac{2v}{1+v^2}\,dv=-\frac{dx}{x}\)

Integrating: \(\log(1+v^2)=-\log x+\log c\Rightarrow x(1+v^2)=c\)

Put \(v=\frac{y}{x}\): \(x\left(1+\frac{y^2}{x^2}\right)=c\Rightarrow x^2+y^2=cx\), which is the general solution.

Doubtly AI · 28 Oct 2026 AI-verified · Oct 2026 Report

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Solve the following differential equation\((x^2-y^2)\,dx+2xy\,dy=0\)

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