2022-01-20

Solve the following first order differential equations:
1. $\left(1+{x}^{2}+{y}^{2}+{x}^{2}{y}^{2}\right)dy={y}^{2}dx$
2. ${e}^{x}y\frac{dy}{dx}={e}^{-y}+{e}^{3x-y}$

### Answer & Explanation

yotaniwc

1. $\left(1+{x}^{2}+{y}^{2}+{x}^{2}{y}^{2}\right)dy={y}^{2}dx$
$\left(\left(1+{x}^{2}\right)+{y}^{2}\left(1+{x}^{2}\right)\right)dy={y}^{2}dx$
$\left(1+{x}^{2}\right)\left(1+{y}^{2}\right)dy={y}^{2}dx$
$\frac{\left(1+{y}^{2}\right)}{{y}^{2}}dy=\frac{dx}{\left(1+{x}^{2}\right)}$
$\int \frac{1}{{y}^{2}}dy+\int dy=\int \frac{1}{1+{x}^{2}}-dx+c$
$\frac{{y}^{-2+1}}{-2+1}+y={\mathrm{tan}}^{-1}x+c$
$-\frac{1}{y}+y={\mathrm{tan}}^{-1}\left(x\right)+c$
$\left(y-\frac{1}{y}\right)={\mathrm{tan}}^{-1}\left(x\right)+c$

2. ${e}^{x}y\frac{dy}{dx}={e}^{-y}+{e}^{3x-y}$

${e}^{x}y\frac{dy}{dx}={e}^{-y}\left(1+{e}^{3x}\right)$
$\frac{ydy}{{e}^{-y}}=\left(\frac{1+{e}^{3x}}{{e}^{x}}\right)dx$
$y{e}^{y}dy=\left({e}^{-x}+{e}^{2x}\right)dx$
$\int y{e}^{y}dy=\int \left({e}^{-x}+{e}^{2x}\right)dx+c$
$y{e}^{y}-{e}^{y}=-{e}^{-x}+\frac{{e}^{2x}}{2}+c$
$2\left(y{e}^{y}-y\right)+2{e}^{x}-{e}^{2x}=c$

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