Question

Solve :
$\frac{dy}{dx}=x(2\log x+1)$, given $y=0$ where $x=2$.

Answer 1

We have,

$\frac{dy}{dx}=x(2{\log }_{e}x+1)$

$dy=\left[x(2{\log }_{e}x+1)\right]dx$

On taking integral both sides, we get

$\int dy=\int \left[x(2{\log }_{e}x+1)\right]dx$

$\int dy=2\int x{\log }_{e}xdx+\int xdx$

$y=2[{\log }_{e}x\left(\frac{x^2}{2}\right)-\int \frac{1}{x}\times \left(\frac{x^2}{2}\right)dx]+\frac{x^2}{2}+C$

$y=2[{\log }_{e}x\left(\frac{x^2}{2}\right)-\frac{1}{2}\int xdx]+\frac{x^2}{2}+C$

$y=2[\left(\frac{x^2}{2}\right){\log }_{e}x-\frac{1}{2}\frac{x^2}{2}]+\frac{x^2}{2}+C$

$y=[x^2{\log }_{e}x-\frac{x^2}{2}]+\frac{x^2}{2}+C$

$y=x^2{\log }_{e}x+C$ ……. (1)

Since, $x=2,y=0$

Therefore,

$0=4{\log }_{e}2+C$

$C=-4{\log }_{e}2$

Therefore,

$y=x^2{\log }_{e}x-4{\log }_{e}2$

Hence, this is the answer.