Solve the differential equation: x x d y d x +y=2 x x - Vidyadip

Question

Solve the differential equation: $\mathrm{x} \log \mathrm{x} \frac{d y}{d x}+\mathrm{y}=\frac{2}{x} \log \mathrm{x}$

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Answer

The given differential equation is
$\mathrm{x} \log \mathrm{x} \frac{d y}{d x}+\mathrm{y}=\frac{2}{x} \log \mathrm{x}$
$\Rightarrow \frac{d y}{d x}+\frac{1}{x \log x} y=\frac{2}{x^{2}}$
This is a linear differential equation of the form
$\frac{d y}{d x}+\mathrm{Py}=\mathrm{Q}$, where $\mathrm{P}=\frac{1}{x \log x}$ and $\mathrm{Q}=\frac{2}{x^{2}}$
$\therefore$ I.F. $=e^{\int P d x}=e^{\int \frac{1}{x \log x} d x}=e^{\int \frac{1}{t} d t}$, where $\mathrm{t}=\log \mathrm{x}$
$\Rightarrow$ I.F. $=\mathrm{e}^{\log \mathrm{t}}=\mathrm{t}=\log \mathrm{x}$
Multiplying both sides of (i) by I.F. $=\log x$, we get
$\log \mathrm{x} \frac{d y}{d x}+\frac{1}{x} \mathrm{y}=\frac{2}{x^{2}} \log \mathrm{x}$
Integrating both sides with respect to $x$, we get
$y \log x=\int \frac{2}{x^{2}} \log \mathrm{xdx}+\mathrm{C}$ [Using: $\mathrm{y}($ I.F. $)=\int \mathrm{Q}$ (I.F.) $\mathrm{dx}+\mathrm{c}$ ]
$\Rightarrow \mathrm{y} \log \mathrm{x}=2 \iint_{I}^{\log } x x_{I I}^{-2} d x+C$
$\Rightarrow \mathrm{y} \log \mathrm{x}=2\left\{\log x\left(\frac{x^{-1}}{-1}\right)-\int \frac{1}{x}\left(\frac{x^{-1}}{-1}\right) d x\right\}+\mathrm{C}$
$\Rightarrow \mathrm{y} \log \mathrm{x}=2\left\{-\frac{\log x}{x}+\int x^{-2} d x\right\}+\mathrm{C}$
$\Rightarrow \mathrm{y} \log \mathrm{x}=2\left\{-\frac{\log x}{x}-\frac{1}{x}\right\}+\mathrm{C}$
$\Rightarrow \mathrm{y} \log \mathrm{x}=-\frac{2}{x}(1+\log \mathrm{x})+\mathrm{C}$, which gives the required solution.

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