Q.24 Solve the differential equation xdy/dx - √(x² + y²) = y

Answer: y + √(x² + y²) = cx² dy/dx = y + √(x² + y²)/x , a homogeneous equation. Put y = vx , dy/dx = v + xdv/dx (taking x gt 0 ): v + xdv/dx = v + √(1 + v²) ⇒ dv/√(1 + v²) = dx/x . Integrating: log|v + √(1 + v²)| = log|x| + log c , so v + √(1 + v²) = cx . Substituting v = y/x : y + √(x² + y²)/x = cx , i.e. y + √(x² + y²) = cx² is the general solution.

Solve the differential equation \(x\frac{dy}{dx} - \sqrt{x^2 + y^2} = y\)

Q.24 Solve the differential equation \(x\frac{dy}{dx} - \sqrt{x^2 + y^2} = y\)

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

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

\(\frac{dy}{dx} = \frac{y + \sqrt{x^2 + y^2}}{x}\), a homogeneous equation. Put \(y = vx\), \(\frac{dy}{dx} = v + x\frac{dv}{dx}\) (taking \(x \gt 0\)):

\(v + x\frac{dv}{dx} = v + \sqrt{1 + v^2} \Rightarrow \frac{dv}{\sqrt{1 + v^2}} = \frac{dx}{x}\).

Integrating: \(\log\left|v + \sqrt{1 + v^2}\right| = \log|x| + \log c\), so \(v + \sqrt{1 + v^2} = cx\).

Substituting \(v = \frac{y}{x}\): \(\frac{y + \sqrt{x^2 + y^2}}{x} = cx\), i.e. \(y + \sqrt{x^2 + y^2} = cx^2\) is the general solution.

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

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Solve the differential equation \(x\frac{dy}{dx} - \sqrt{x^2 + y^2} = y\)

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