Question

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

• A. $2$
• B. $2e$
• C. $e$
• D. $0$

Answer 1

The correct option is A $2$

$\left(x\log x\right)\frac{dy}{dx}+y=2x\log x$

$\to$ dividing by $x\log x$ , we get

$\to \frac{dy}{dx}+\frac{y}{x\log x}=2$

Multiplying by IF, we get

$\to d(IF\times y)=2\log xdx$

$\to IF=e^{\int \frac{1}{x\log x}dx}=e^{\log \left(\log x\right)}=\log x$

$\rightarrow Multiplying by IF, we get

$\to d(IF\times y)=2\log xdx$

By integrating, we get

$\to y\log x=\int 2\log xdx$

By using Product Rule on RHS, we get

$\to y\log x=2[\log x\int 1-\int ((\int 1)\frac{d}{dx}\left(\log x\right))dx$

$\to y\log x=2\left[x\right(\log x-1\left)\right]+C$

Put $x=1$, we get

$0=2\left[1\right(0-1\left)\right]+C$

$C=2$

$\to y\log x=2\left[x\right(\log x-1\left)\right]+2$

at $x=e$

$y=2$