Question

Let $y\left(x\right)$ be the solution of the differential equation $\left(x\log x\right)\frac{dy}{dx}+y=2x\log x,(x⩾1)$. Then $y\left(e\right)$ is equal to :

• A. $e$
• B. $0$
• C. $2$
• D. $2e$

Answer 1

Answer: (C)

$2$

Given differential equation may be written as,
$\frac{dy}{dx}+\frac{y}{x\log x}=2$
Clearly this is of the form, $\frac{dy}{dx}+Py=Q$
So, I.F. $=e^{\int pdx}=e^{\int \frac{1}{x\log x}dx}=e^{ln\left(lnx\right)}=\ln x$
$\Rightarrow y(I.F.)=\int Q\left(x\right)\times I.F.dx$
$\Rightarrow y\left(lnx\right)=\int 2\times \ln xdx$
$\Rightarrow y\left(lnx\right)=2(x\ln x-x)+c$
Put $x=1$ we get, $c=2$
$y\left(e\right)-0=2\Rightarrow y\left(e\right)=2$