Question

Let the solution curve of the differential equation $xdy=(\sqrt{x^2+y^2}+y)dx,x>0$, intersect the line $x=1$ at $y=0$ and the line $x=2$ at $y=\alpha$. Then the value of $\alpha$ is

• A. $-\frac{3}{2}$
• B. $\frac{3}{2}$
• C. $\frac{1}{2}$
• D. $\frac{5}{2}$

Answer 1

The correct option is B $\frac{3}{2}$

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

$\Rightarrow \frac{dy}{dx}=\sqrt{1+\frac{y^2}{x^2}}+\frac{y}{x}$

Let $\frac{y}{x}=v$

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

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

OR $\ln (v+\sqrt{1+v^2})=\ln x+C$

at $x=1,y=0$

$\Rightarrow C=0$

$\frac{y}{x}+\sqrt{1+\frac{y^2}{x^2}}=x$

At $x=2$

$\frac{y}{2}+\sqrt{1+\frac{y^2}{4}}=2$

$\Rightarrow 1+\frac{y^2}{4}=4+\frac{y^2}{4}-2y$

OR $y=\frac{3}{2}$