Question

In Problems 1–22 solve the given differential equation by separationof variables.${\mathbf{\text{e}}}^{\mathbf{\text{x}}}\mathbf{\text{y}}\frac{\mathbf{\text{dy}}}{\mathbf{\text{dx}}}{\mathbf{\text{=e}}}^{\mathbf{\text{-y}}}{\mathbf{\text{+e}}}^{\mathbf{\text{-2x-y}}}$

Short answer

The solution of the given differential equation is$e^y(y-1)=-e^{-x}-\frac{1}{3}{e}^{-3x}+C$.

Step-by-step solution

Step 1:Separable Equation

A first-order differential equation of the form $\frac{dy}{dx}=g\left(x\right)h\left(y\right)$is said to be separable or to have separable variables.

Separate the variables and integrate

The given equation is$e^{x}y\frac{dy}{dx}=e^{-y}+e^{-2x-y}$,.

Separate the variables,

$y\frac{dy}{e^{-y}}=\frac{(1+e^{-2x})}{e^x}dx$

Apply,$\left(\frac{1}{e^n}=e^{-n}\right)$we get

$y{e}^{y}dy=(e^{-x}+e^{-3x})dx$

Integrate both sides,

$\int y{e}^{y}dy=\int e^{-x}dx+\int e^{-3x}dx$

Integrating by parts,$\int y{e}^{y}dy$

$\begin{array}{c}u=v\Rightarrow du=dy\\ dv=e^{y}dy\Rightarrow v=e^y\end{array}$

So,

$\begin{array}{l}\int y{e}^{y}dy=y{e}^y-\int e^{y}dy\\ \int y{e}^{y}dy=y{e}^y-e^y+C\end{array}$

For$\int e^{-x}dx\text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}}\int e^{-3x}dx$

Apply role="math" $\int e^{\pm ax}dx=\pm \frac{1}{a}{e}^{-ax}+C$So,

$y{e}^y-e^y=-e^{-x}-\frac{1}{3}{e}^{-3x}+C$

Simplify,

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

Hence, the solution of the given differential equation is.${\mathbf{\text{e}}}^{\mathbf{\text{y}}}{\mathbf{\text{(y-1)=-e}}}^{\mathbf{\text{-x}}}\mathbf{\text{-}}\frac{\mathbf{\text{1}}}{\mathbf{\text{3}}}{\mathbf{\text{e}}}^{\mathbf{\text{-3x}}}\mathbf{\text{+C}}$