Question

If $x^2\frac{dy}{dx}+2xy=\frac{2ln\left(x\right)}{x},$ and $y\left(1\right)=0$, then what is $y\left(e\right)$?

• A. $e$
• B. $1$
• C. $1/e$
• D. $1/e^2$

Answer 1

The correct option is D $1/e^2$

Given the differential equation
$x^2\frac{dy}{dx}+2xy=\frac{2logx}{x},y\left(1\right)=0$
$\Rightarrow \frac{dy}{dx}+\left(\frac{2}{x}\right)y=\frac{2logx}{x^3}$ ...... (i)
$IF=e^{\int \frac{2}{x}dx}=e^{2lnx}=x^2$
Solution of (i) is
$y.x^2=\int \frac{2logx}{x^3}{x}^{2}dx+c$
$=\int \frac{2logx}{x}dx+c$
$y.x^2=\frac{2(logx{)}^2}{2}+c$
$y={\left(\frac{logx}{x}\right)}^2+\frac{c}{x^2}$
$y\left(1\right)=0\Rightarrow +{\left(\frac{log1}{1}\right)}^2+c=0$
$\Rightarrow c=0$
$y\left(x\right)={\left(\frac{logx}{x}\right)}^2$
$y\left(e\right)=\frac{1}{e^2}$