Solve y d x-x d y=- x d x - Vidyadip

Question

Solve $y d x-x d y=-\log x d x$

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Answer

$
\begin{aligned}
& y d x-x d y=-\log x d x \\
& \therefore ydx - xdy +\log xdx =0 \\
& \therefore xdy =( y +\log x ) dx \\
& \therefore \frac{d y}{d x}=\frac{y+\log x}{x}=\frac{y}{x}+\frac{\log x}{x} \\
& \therefore \frac{d y}{d x}-\frac{1}{x} \cdot y=\frac{\log x}{x} \\
\end{aligned}
$
This is the linear differential equation of the form
$
\begin{aligned}
& \frac{d y}{d x}+P \cdot y=Q, \text { where } P=-\frac{1}{x} \text { and } Q=\frac{\log x}{x} \\
& \begin{aligned}
\therefore \text { I.F. } & =e^{\int P d x}=e^{\int-\frac{1}{x} d x}=e^{-\log x} \\
& =e^{\log (x)^{-1}}=\frac{1}{x}
\end{aligned}
\end{aligned}
$
$\therefore$ the solution of (1) is given by
$
\begin{aligned}
& y \cdot\left(\text { I.F.) }=\int Q \cdot(\text { I.F.) } d x+c\right. \\
\therefore & y \cdot \frac{1}{x}=\int \frac{\log x}{x} \cdot \frac{1}{x} d x+c \\
\therefore & \frac{y}{x}=\int \frac{\log x}{x^2} d x+c \\
\therefore & \frac{y}{x}=(\log x) \int x^{-2} d x-\int\left[\frac{d}{d x}(\log x) \int x^{-2} d x\right] d x+c \\
\therefore & \frac{y}{x}=(\log x) \cdot \frac{x^{-1}}{-1}-\int \frac{1}{x} \cdot \frac{x^{-1}}{-1} d x+c \\
\therefore & \frac{y}{x}=-\frac{\log x}{x}+\int x^{-2} d x+c \\
\therefore & \frac{y}{x}=-\frac{\log x}{x}+\frac{x^{-1}}{-1}+c \\
\therefore & \frac{y}{x}=-\frac{\log x}{x}-\frac{1}{x}+c
\end{aligned}
$
$
\begin{aligned}
& \therefore y=-\log x-1+c x \\
& \therefore y=c x-(1+\log x)
\end{aligned}
$This is the general solution.

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