Question

The solution of the differential equation

xdy – y dx = $\sqrt{x^{2} + y^{2}}$dx is –

• A. x + $\sqrt{x^{2} + y^{2}}$ = cx²
• B. y – $\sqrt{x^{2} + y^{2}}$ = cx
• C. x – $\sqrt{x^{2} + y^{2}}$= cx
• D. y + $\sqrt{x^{2} + y^{2}}$= cx²

Answer 1

Answer: (D)

y + $\sqrt{x^{2} + y^{2}}$= cx²

Answer 2

Given that, x dy – y dx =$\sqrt{x^{2} + y^{2}}$dx

Ž xdy = ($\sqrt{x^{2} + y^{2}}$ + y) dx

Ž $\frac{dy}{dx}$= $\frac{\sqrt{x^{2} + y^{2}} + y}{x}$

Now, put y = vx and $\frac{dy}{dx}$ = v + x $\frac{dv}{dx}$

\ v + x $\frac{dv}{dx}$ = $\frac{\sqrt{x^{2} + v^{2}x^{2}} + vx}{x}$

Ž x $\frac{dv}{dx}$ = $\sqrt{1 + v^{2}}$ + v – v = $\sqrt{1 + v^{2}}$

On integrating both sides

$\int_{}^{}\frac{dv}{\sqrt{1 + v^{2}}}$ = $\int_{}^{}\frac{dx}{x}$

Ž log (v +$\sqrt{1 + v^{2}}$) = log x + log c

Ž y + $\sqrt{x^{2} + y^{2}}$= cx²