Question

A curve passes thorugh the point $(x=1,y=0)$ and satisfies the differential equation $\frac{dy}{dx}=\frac{x^2+y^2}{2y}+\frac{y}{x}$
The equation that decribes the curve is

• A. $ln(1+\frac{y^2}{x^2})=x-1$
• B. $\frac{1}{2}ln(1+\frac{y^2}{x^2})=x-1$
• C. $ln(1+\frac{y}{x})=x-1$
• D. $\frac{1}{2}ln(1+\frac{y}{x})=x-1$

Answer 1

The correct option is A $ln(1+\frac{y^2}{x^2})=x-1$

$\frac{dy}{dx}=\frac{x^2+y^2}{2y}+\frac{y}{x}$ ..... (1)
Put $y=vx$
$\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
$v+x\frac{dv}{dx}=\frac{x^2+v^2{x}^2}{2vx}+v$
$x\frac{dv}{dx}=\frac{x}{2v}+\frac{vx}{2}$
$\Rightarrow \frac{dv}{dx}=\frac{1}{2}[\frac{1}{v}+v]$
$\left(\frac{2v}{1+v^2}\right)dv=dx$
$\Rightarrow log(1+v^2)=v+c$
$log[1+\frac{y^2}{x^2}]=x+c$
Now, $y\left(1\right)=0$
$\Rightarrow C=-1$
so $log(1+\frac{y^2}{x^2})=x-1$