Solve y y d x d y + x - y =0 - Vidyadip

Question

Solve y $\log y \frac{d x}{d y}+ x -\log y =0$

✓

Answer

$
\begin{aligned}
& y \log y \cdot \frac{d x}{d y}+x-\log y=0 \\
& \therefore y \log y \cdot \frac{d x}{d y}=\log y-x \\
& \therefore \frac{d x}{d y}=\frac{\log y-x}{y \log y} \\
& \therefore \frac{d x}{d y}=\frac{1}{y}-\frac{x}{y \log y} \\
& \therefore \frac{d x}{d y}+\frac{x}{y \log y}=\frac{1}{y}
\end{aligned}
$
This is the linear differential equation of the form $\frac{d x}{d y}+P x=Q$, where $P=\frac{1}{y \log y}$ and $Q=\frac{1}{y}$
$
\begin{aligned}
\therefore \text { I.F. } & =e^{\int P d y}=e^{\int \frac{1}{y \log y} d y} \\
& =e^{\int \frac{(1 / y)}{\log y} d y}=e^{\log |\log y|}=\log y
\end{aligned}
$
$\therefore$ the solution of (1) is given by
$
\begin{aligned}
& x \cdot(\text { I.F. })=\int Q \cdot(\text { I.F. }) d y+c_1 \\
\therefore x \cdot \log y & =\int \frac{1}{y} \cdot \log y d y+c_1 \\
\therefore \quad & (\log y) \cdot x=\int \frac{\log y}{y} d y+c_1
\end{aligned}
$
Put $\log y=t \quad \therefore \frac{1}{y} d y=d t$
$
\begin{aligned}
& \therefore(\log y) \cdot x=\int t d t+c_1 \\
& \therefore x \log y=\frac{t^2}{2}+c_1
\end{aligned}
$
$
\begin{aligned}
& \therefore x \log y=\frac{1}{2}(\log y)^2+c_1 \\
& \therefore 2 x \log y=(\log y)^2+c, \text { where } c=2 c_1
\end{aligned}
$
This is the general solution.

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