Question

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

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

Answer 1

Answer: (B)

$xdy=\left(\sqrt{x^2+y^2}+y\right)dx$
$xdy-ydx=\sqrt{x^2+y^2}dx$
$\frac{xdy-ydx}{x^2}=\sqrt{1+\frac{y^2}{x^2}}\cdot \frac{dx}{x}$
$\frac{d(y/x)}{\sqrt{1+{\left(\frac{y}{x}\right)}^2}}=\frac{dx}{x}$
$\ln \left(\frac{y}{x}+\sqrt{{\left(\frac{y}{x}\right)}^2+1}\right)=\ln x+R$
$\frac{y+\sqrt{y^2+x^2}}{x}=cx$
$y+\sqrt{y^2+x^2}=c{x}^2$
$x=1,y=0\Rightarrow 0+1=C\Rightarrow C=1$
Curve is $y+\sqrt{x^2+y^2}=x^2$
$x=2,y=\alpha$
$2+\sqrt{4+{\alpha }^2}=4$
$4+{\alpha }^2=16+{\alpha }^2=8\alpha$
$\alpha =\frac{3}{2}$