Question

The solution of the differential equation \[ x\,dy - y\,dx = \sqrt{x^2 + y^2}\,dx \] (where \(c\) is the constant of integration) is

• A. \( \sqrt{x^2+y^2} = cx^2 - y \)
• B. \( \sqrt{x^2+y^2} = cx^2 + y \)
• C. \( \sqrt{x^2+y^2} = cx - y \)
• D. \( \sqrt{x^2+y^2} = cx + y \)

Hint:

Whenever you see \(x\,dy - y\,dx\), immediately think of \[ d\!\left(\frac{y}{x}\right)=\frac{x\,dy-y\,dx}{x^2} \] This identity simplifies many differential equations quickly.

Answer 1

The Correct Option is D

Concept: Expressions of the form \(x\,dy - y\,dx\) are commonly simplified using the identity: \[ d\!\left(\frac{y}{x}\right)=\frac{x\,dy-y\,dx}{x^2} \] Such differential equations are generally solved by converting them into functions of \(\dfrac{y}{x}\) or by suitable substitutions involving \(\sqrt{x^2+y^2}\).
Step 1: Rewrite the given equation. \[ x\,dy - y\,dx = \sqrt{x^2+y^2}\,dx \] Divide both sides by \(x^2\): \[ \frac{x\,dy-y\,dx}{x^2}=\frac{\sqrt{x^2+y^2}}{x^2}\,dx \]
Step 2: Use the differential identity. \[ d\!\left(\frac{y}{x}\right)=\frac{x\,dy-y\,dx}{x^2} \] Thus, \[ d\!\left(\frac{y}{x}\right)=\frac{\sqrt{x^2+y^2}}{x^2}\,dx \]
Step 3: Simplify the right-hand side. \[ \frac{\sqrt{x^2+y^2}}{x^2} = \frac{\sqrt{x^2\left(1+\left(\frac{y}{x}\right)^2\right)}}{x^2} = \frac{1}{x}\sqrt{1+\left(\frac{y}{x}\right)^2} \] Let \[ u=\frac{y}{x} \] Then, \[ du=\frac{1}{x}\sqrt{1+u^2}\,dx \]
Step 4: Integrate both sides. \[ \int du = \int \frac{1}{x}\sqrt{1+u^2}\,dx \] This gives: \[ u = \sqrt{1+u^2}+c \] Substituting back \(u=\dfrac{y}{x}\): \[ \frac{y}{x} = \frac{\sqrt{x^2+y^2}}{x}+c \] Multiplying throughout by \(x\): \[ y = \sqrt{x^2+y^2} + cx \] Rearranging, \[ \sqrt{x^2+y^2} = cx + y \]