Question

Let $y^{\prime}=\frac{4 y^2+4 x y+x^2}{4 x^2}$ and $y(1)=0$, then $y\left(e^{\frac{\pi}{2}}\right)$ equals

• A. $\frac{1}{2} e^{\frac{\pi}{2}}$
• B. $e ^{\frac{\pi}{2}}$
• C. $\frac{1}{4} e ^{\frac{\pi}{2}}$
• D. $\frac{\pi}{2}$

Answer 1

Answer: (A)

Put $y=v x$
$v+\frac{x d v}{d x}=\frac{4 v^2+4 v+1}{4} \Rightarrow \frac{x d v}{d x}=\frac{4 v^2+1}{4} $
$\int \frac{d v}{v^2+\frac{1}{4}}=\ln x+C $
$2 \tan ^{-1}(2 v)=\ln x+C $
$x=1, y=0 $
$\therefore \quad C=0 $
$2 v=\tan \left(\frac{1}{2} \ln x\right)$
$y=\frac{x}{2} \tan \left(\frac{1}{2} \ln x\right)$
$y\left(e^{\pi / 2}\right)=\frac{e^{\pi / 2}}{2} $