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

Question

$y \log y \frac{d x}{d y}=\log y - x$

✓

Answer

$y \log y \frac{d x}{d y}=\log y-x$
$y \log y \frac{d x}{d y}+x=\log y$
$\therefore \frac{d x}{d y}+\frac{1}{y \log y} x=\frac{1}{y}$
The given equation is of the form $\frac{d x}{d y}+p x=Q$
where, $P=\frac{1}{y \log y}$ and $Q=\frac{1}{y}$
$\therefore I . F .=e^{\int^{p d y}=e^{\int^{\frac{1}{y \log y}} d y}=e^{\log \mid \log y}=\log y}$
$\therefore$ Solution of the given equation is
$x(I . F)=.\int Q(I . F) d y+.c_1$
$\therefore x \cdot \log y=1 y \int \log y d y+c_1$
In R. H. S., put $\log y=t$
Differentiating w.r.t. $x$, we get
$
\frac{1}{y} d y=d t
$
$\begin{aligned} & \therefore x \log y=t d t \int+c_1=\frac{t^2}{2}+c_1 \\ & \therefore x \log y=\frac{(\log y)^2}{2}+c_1 \\ & \therefore 2 x \log y=(\log y)^2+c \ldots\left[2 c_1=c\right] \\ & x \log y=\frac{1}{2}(\log y)^2+c\end{aligned}$

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