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Solve the equation $x^{2} e^{y}\left(\frac{d x}{d y}+x\right)=1$

$x^{2} e^{y}\left(\frac{d x}{d y}+x\right)=1 \quad * x^{-2} e^{-y}$

$\left(\frac{d x}{d y}+x\right)=x^{-2} e^{-y} \quad * x^{2}$

$x^{2}\left(\frac{d x}{d y}+x\right)=e^{-y} \longrightarrow (1)$

$x^{2} \frac{d x}{d y}+x^{3}=e^{-y} \cdots (1)$

let $z=x^{3} \rightarrow \frac{d z}{d y}=3 x^{2} \frac{d x}{d y} \quad \div 3$

$\frac{1}{3} \frac{d z}{d y}=x^{2} \frac{d x}{d y}$ substitute in $(1)$

$\frac{1}{3} \frac{d z}{d y}+z=e^{-y} \quad * 3$

$\frac{d z}{d y}+3 z=3 e^{-y}$ Linear in $z$

$\mu=e^{\int {3} d y}=e^{3 y}$

$\mu {z}=\int \mu \cdot Q(y) d y+c$

$e^{3 y} z=\int e^{3 y} \cdot 3 e^{-y} d y+c$

$\longrightarrow e^{3 y} x^{3}=\int 3 e^{2 y}+c$

$e^{3 y} x^{3}=3 \frac{1}{2} e^{2 y}+c$

$\longrightarrow e^{3 y} x^{3}=\frac{3}{2} e^{2 y}+c$

Solve the equation $y(x \tan x+\ln y) \mathrm{d} x+\tan x \mathrm{d} y=0$

$y(x \tan x+\ln y) d x+\tan x d y=0 \quad \div y$

$(x \tan x+\ln y) d x+ \frac{\tan x d y}{y}=0$

let $u=\ln y \rightarrow d u=\frac{1}{y} d y$

$(x \tan x+u) d x+\tan x d u=0 \quad \div d x$

$\tan x \frac{d u}{d x}+u=-x \tan x \quad \div \tan x$

$\frac{d u}{d x}+\frac{u}{\tan x}=-x$

$\frac{1}{\tan x}=\cot x$

$\longrightarrow \frac{d u}{d x}+\cot x u=-x$

Linear in $u$

$\frac{d u}{d x}+(\cot x) u=-x$

$\mu=e^{\int \cot x d x}=e^{\ln |\sin x|}=\sin x$

$\mu \cdot u=\int \mu Q(x) d x+C$

$\rightarrow \sin (x) \cdot u=\int \sin (x)(-x) d x+c$

by parts

$u=-x \quad d u=-1 d x$
$v=-\cos x \quad d u=\sin (x)$

$\sin x \cdot u=x \cos x-\int-\cos x(-1) d x + c$

$\sin x \cdot u=x \cos x-\sin x+c$