zagonek34

2021-12-27

Using variable separable find the complete solution of the given differential equations
$x{y}^{3}dx+\left(y+1\right){e}^{-x}dy$

Maria Lopez

$x{y}^{3}dx+\left(y+1\right){e}^{-x}dy=0$
$x{y}^{3}dx=-\left(y+1\right){e}^{-x}dy$
$\frac{xdx}{{e}^{-x}}=\frac{-\left(y+1\right)}{{y}^{3}}dy$
$x{e}^{x}dx=\frac{-y}{{y}^{3}}-\frac{1}{{y}^{3}}dy$
$x{e}^{x}dx=-{y}^{-2}-{y}^{-3}dy$
Integrate on both sides
$\int x{e}^{x}dx=\int -{y}^{-2}-{y}^{-3}dy$
$x{e}^{x}-{e}^{x}=\frac{-{y}^{-2+1}}{-2+1}-\frac{{y}^{-3+1}}{-3+1}+c$
$x{e}^{x}-{e}^{x}=\frac{1}{y}+\frac{1}{2{y}^{2}}+c$

Travis Hicks

Step 1
The given equation is:
$x{y}^{3}dx+\left(y+1\right){e}^{-x}dy=0$
We can rewrite this as:
$x{y}^{3}dx=-\left(y+1\right){e}^{-x}dy$
Divide both sides by ${e}^{-x}{y}^{3}$
$⇒x{e}^{x}dx=-\left(y+1\right){y}^{-3}dy$
This is variable separable form
Step 2
We can solve the equation by integrating equation we have got in Step 1
$⇒\int x{e}^{x}dx=\int -\left(y+1\right){y}^{-3}dy$
$⇒x{e}^{x}-\int {e}^{x}.1dx=-\int \left({y}^{-2}+{y}^{-3}\right)dy$
$⇒x{e}^{x}-{e}^{x}=-\frac{{y}^{-2+1}}{-2+1}-\frac{{y}^{-3+1}}{-3+1}+C$
$⇒{e}^{x}\left(x-1\right)=\frac{1}{y}+\frac{1}{2{y}^{2}}+C$
$⇒{e}^{x}\left(x-1\right)=\frac{2y+1}{2{y}^{2}}+C$
Step 3
Answer: ${e}^{x}\left(x-1\right)=\frac{2y+1}{2{y}^{2}}+C$

karton

$\begin{array}{}x{y}^{3}dx+\left(y+1\right){e}^{-x}dy=0\\ x{y}^{3}dx=-\left(y+1\right){e}^{-x}dy\\ \frac{xdx}{{e}^{-x}}=\frac{-\left(y+1\right)}{{y}^{3}}dy\\ x{e}^{x}dx=\frac{-y}{{y}^{3}}-\frac{1}{{y}^{3}}dy\\ x{e}^{x}dx=-{y}^{-2}-{y}^{-3}dy\\ \int x{e}^{x}dx=\int -{y}^{-2}-{y}^{-3}dy\\ x{e}^{x}-{e}^{x}=\frac{1}{y}+\frac{1}{2{y}^{2}}+c\end{array}$