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 involving \(x\,dy - y\,dx\) can be simplified using the identity: \[ d\!\left(\frac{y}{x}\right)=\frac{x\,dy-y\,dx}{x^2} \] Such differential equations are usually solved by rewriting them in terms of \(\dfrac{y}{x}\) or by making appropriate substitutions involving \(\sqrt{x^2+y^2}\).
Step 1: Rewrite the given differential 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: Apply the differential identity. \[ d\!\left(\frac{y}{x}\right)=\frac{x\,dy-y\,dx}{x^2} \] Hence, \[ d\!\left(\frac{y}{x}\right)=\frac{\sqrt{x^2+y^2}}{x^2}\,dx \]
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 the equation becomes: \[ du=\frac{1}{x}\sqrt{1+u^2}\,dx \]
Step 4: Integrate both sides. \[ \int du = \int \frac{1}{x}\sqrt{1+u^2}\,dx \] This yields: \[ u = \sqrt{1+u^2}+c \] 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 the terms: \[ \sqrt{x^2+y^2} = cx + y \]