Question

Find the general solution of the differential equation $xlogx\frac{dy}{dx}+y=\frac{2}{x}logx$.

Answer 1

$xlogx\frac{dy}{dx}+y=\frac{2}{x}logx$

Put in form $\frac{dy}{dx}+Py=Q$

$xlogx\frac{dy}{dx}+y=\frac{2}{x}logx$

$\frac{dy}{dx}+\left(\frac{1}{xlogx}\right)y=\frac{2}{x^2}$ .....(1)

Therefore,

$P=\frac{1}{xlogx}$ and $Q=\frac{2}{x^2}$

Hence, Integrating factor,

IF = $e^{\int Pdx}=e^{\int \frac{1}{xlogx}dx}$

Let $t=logx$

$dt=\frac{1}{x}dx$

$IF=e^{\int \frac{1}{t}dt}$

$IF=e^{logt}=t=logx$

Multiplying equation 1 by IF, we get,

$ylogx=2\int logx{x}^{-2}dx$ ......(2)

Let $I=2\int x^{-2}logxdx$

$I=2[logx\int x^{-2}dx-\int \frac{1}{x}[\int x^{-2}dx\left]dx\right]$

$=2[-logx\frac{1}{x}+\int \frac{1}{x^2}dx]$

$=-2[\frac{-1}{x}logx-\frac{1}{x}]$

$=\frac{-2}{x}(1+logx)$

Thus, the equation 2 becomes,

$ylog|x|=\frac{-2}{x}(1+log|x|)+C$ which is the general solution of the given equation.