Question

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

Answer 1

Given, $xlogx\frac{dy}{dx}+y=\frac{2}{x}logx$
On dividing by x~logx, on both sides, we get $\frac{dy}{dx}+\frac{y}{xlogx}=\frac{2}{x^2}$
On comparing with the form $\frac{dy}{dx}+Py=Q,$ we get $P=\frac{1}{xlogx}$ and $Q=\frac{2}{x^2}$
$\therefore IF=e^{\int pdx}=e^{\int \frac{1dx}{xlogx}}Letlogx=t\Rightarrow \frac{1}{x}=\frac{dt}{dx}\Rightarrow dx=xdt\therefore IF=e^{\int \frac{1}{t}dt}=e^{log|t|}=t=logx$ $[\because t=logx]$
The general solution of the given differential equation is given by,
$y.IF=\int QIFdx+C\Rightarrow y.logx=\int \frac{2}{x^2}logxdx+C\Rightarrow ylogx=2\int (logx.\frac{1}{x^2})dx+C=2[logx\int \frac{1}{x^2}dx-\int \left(\frac{d}{dx}\right(logx)\int \frac{1}{x^2}dx)dx]+C=2\left[logx(-\frac{1}{x}-\int \{\frac{1}{x}.(-\frac{1}{x})\}dx)\right]+C=2[-\frac{logx}{x}+\int \frac{1}{x^2}dx]+C=2[-\frac{logx}{x}-\frac{1}{x}]+C\Rightarrow ylogx=-\frac{2}{x}(1+logx)+C$