Solve the following differential equation y y d x d y +x= y - Vidyadip

Question

Solve the following differential equation
$y \log y \frac{ d x}{ d y}+x=\log y$

✓

Answer

$ y \log y \frac{ d x}{ d y}+x=\log y$
$\therefore \frac{ d x}{ d y}+\frac{1}{y \log y} x=\frac{1}{y} $
The given equation is of the form $\frac{ d x}{ d y}+ P x= Q$ where, $P =\frac{1}{y \log y}$ and $Q =\frac{1}{y}$
$ \therefore \text { I.F. }= e ^{\int^{ Pd y}}$
$= e ^{\int^{\frac{1}{\log y} d y}}$
$= e ^{\log |\log y|}$
$=\log y $
$\therefore$ Solution of the given equation is
$ x ( I.F .)=\int Q ( I . F .) d y+ c _1$
$\therefore x . \log y =\int \frac{1}{y} \log y d y+ c _1 $
In R. H. S., put $\log y=t$
Differentiating w.r.t. $x$, we get
$\frac{1}{y} d y= dt$
$\therefore x \log y =\int t dt + c _1$
$=\frac{ t ^2}{2}+ c _1$
$\therefore x \log y =\frac{(\log y)^2}{2}+ c _1$
$\therefore 2 x \log y =(\log y )^2+ c \quad \ldots . . .\left[2 c _1+ c \right]$

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