David Troyer

2021-12-26

For each of the following differential equations, determine the general or particular solution: $5x{e}^{-y}+2\mathrm{cos}\left(3x\right)\right]{y}^{\prime }+5{e}^{-y}-3\mathrm{sin}\left(3x\right)=0$

Determine the general or particular
$5x{e}^{-y}+2\mathrm{cos}\left(3x\right)\right]{y}^{\prime }+5{e}^{-y}-3\mathrm{sin}\left(3x\right)=0$
$\left(5x{e}^{-y}+2\mathrm{cos}3x\right)dy+\left(5{e}^{-y}-3\mathrm{sin}3x\right)dx=0$
$⇒\left(5{e}^{-y}-3\mathrm{sin}3x\right)dx+\left(5x{e}^{-y}+2\mathrm{cos}3x\right)dy=0$
It is of form $Mdx+Ndy=0$
$\frac{\partial M}{\partial y}=-5{e}^{-y}$
$\frac{\partial N}{\partial x}=5{e}^{-y}-6\mathrm{sin}3x$
Now, $\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}=10{e}^{-y}-6\mathrm{sin}3x$
$\frac{\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}}{M}=\frac{2\left(5{e}^{-y}-3\mathrm{sin}3x\right)}{\left(5{e}^{-y}-3\mathrm{sin}3x\right)}=2$
$I.F={e}^{\int 2.dy}={e}^{2y}$
Multiply eq (i) by ${e}^{2y}$ both sides
$⇒\left(5{e}^{y}-3{e}^{2y}\cdot \mathrm{sin}3x\right)dx+\left(5x{e}^{y}+2{e}^{2y}\cdot \mathrm{cos}3x\right)dy=0$
Now above differential equation is exact
$⇒\left(5{e}^{y}dx+5x{e}^{y}dy\right)+\left(2{e}^{2y}\cdot \mathrm{cos}3x\cdot dy-3{e}^{2y}\cdot \mathrm{sin}3x\right)=0$

godsrvnt0706

Given differential equation $\left({e}^{x}+1\right)\frac{dy}{dx}=y-y{e}^{x}$
$\left({e}^{x}+1\right)\frac{dy}{dx}=y\left(1-{e}^{x}\right)$
on separaing the variables, we get
$\frac{dy}{y}=\frac{1-{e}^{x}}{1+{e}^{x}}dx$
integrating on both sides
$\int \frac{dy}{y}=\int \frac{1-{e}^{x}}{1+{e}^{x}}dx$
$\mathrm{ln}y=\int \frac{\left(1+{e}^{x}\right)-2{e}^{x}}{1+{e}^{x}}dx$
$\mathrm{ln}y=\int \left(1-2\frac{{e}^{x}}{1+{e}^{x}}\right)dx$
$\mathrm{ln}y=x-2\mathrm{ln}\left(1+{e}^{x}\right)+c$
$y={e}^{x-2\mathrm{ln}\left(1+e\right)+c}$
$y={e}^{x}{e}^{-2\mathrm{ln}\left(1+e\right)}{e}^{c}$

$y={c}_{1}{e}^{x}{\left(1+{e}^{x}\right)}^{-2}$
Therefore the solution of the given differential equation is $y={c}_{1}\frac{{e}^{x}}{{\left(1+{e}^{x}\right)}^{2}}$

karton