Question

Find the general solution of the differential equation $x\text{log}x\frac{dy}{dx}+y=\frac{2}{x}\text{log}x$

Answer 1

Given differential equation :

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

$\frac{dy}{dx}+\frac{y}{x\text{log}x}=\frac{2}{x}\text{log}x\times \frac{1}{x\text{log}x}$

$\frac{dy}{dx}+\left(\frac{1}{x\text{log}x}\right)=\frac{2}{x^2}$

Given differential equation is of the form
$\frac{dy}{dx}+Py=Q$

By comparing both the equations, we get

$P=\left(\frac{1}{x\text{log}x}\right)$ and $Q=\frac{2}{x^2}$

The general solution of the given differential equation is

$y(I.F)=\int (Q\times I.F.)dx+c$ ....(i)

Firstly, we need to find I.F.

$I.F.=e^{\int pdx}$

$I.F.=e^{\int \frac{1}{x\text{log}x}dx}$

Let $\text{log}x=t$
$dx=xdt$

$I.F.=e^{\int \frac{1}{xt}\times xdt}$

$I.F.=e^{\int \frac{1}{t}dt}$

$I.F.=e^{\text{log}|t|}$

$I.F.=|t|$

Substituting $t=\text{log}x$, we get

$I.F.=\text{log}x$

Substituting the value of I.F in (i), we get

$y\text{log}x=\int \frac{2}{x^2}.\text{log}x.dx+c$........(ii)

$y\text{log}x=2\int \text{log}x.x^{-2}dx+c$

Using Integration by parts :

$\int f\left(x\right)g\left(x\right)dx=f\left(x\right)\int g\left(x\right)dx-\int \left[f^{\prime }\right(x)\int g(x\left)dx\right]dx$

Taking $f\left(x\right)=\text{log}x$ and $g\left(x\right)=x^{-2}$

$y\text{log}x=2[\text{log}x\int x^{-2}dx-\int \frac{1}{x}[x^{-2}dx\left]dx\right]$

$y\text{log}x=2[\text{log}x\frac{x^{-1}}{(-1)}-\int \frac{1}{x}.\frac{\left(x^{-1}\right)}{(-1)}dx]$

$y\text{log}x=2[\frac{-\text{log}x}{x}+\int \frac{1}{x^2}.dx]$

$y\text{log}x=2[\frac{-\text{log}x}{x}-\frac{1}{x}]$

$\therefore y\text{log}y=\frac{-2}{x}(1+\text{log}x-1)+c$