Question

In Problems 21-26 solve the given initial-value problem.

$\left(e^x+y\right)\mathit{d}\mathit{x}\mathbf{+}\left(2+x+y{e}^y\right)\mathit{d}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$

Short answer

The initial-value of the problem is$2y+xy+y{e}^y-e^y+e^x=3$

Step-by-step solution

Given Information

The given equation is, $\left(e^x+y\right)dx+\left(2+x+y{e}^y\right)dy=0,y\left(1\right)=1$.

Compare the given equation with $Mdx+Ndy=0$

$\mathit{M}\mathbf{(}\mathit{x}\mathbf{,}\mathit{y}\mathbf{)}\mathbf{=}{\mathit{e}}^{\mathbf{x}}\mathbf{+}\mathit{y}$

$N(x,y)=2+x+y{e}^y$

Condition for exactness

An equation of the form $Mdx+Ndy=0$, is said to be exact, if $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$.

Determining the exactness of the differential equation

Checking whether the differential equation is exact by finding $\frac{\partial M}{\partial y},\frac{\partial N}{\partial x}$.

$$\begin{gathered} \frac{\partial M}{\partial y}=1 \\ \frac{{\displaystyle \partial N}}{{\displaystyle \partial x}}=1 \end{gathered}$$

As a result, the equation is exact,$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$

Find the solution

Now, find the solution of the given equation, where $f(x,y)=2y+xy+y{e}^y-e^y+g\left(x\right)$.

Integrate $f(x,y)=2y+xy+y{e}^y-e^y+g\left(x\right)$

$\frac{\partial f}{\partial x}=y+g^{¢}\left(x\right)$

Take $\frac{\partial f}{\partial x}$

$g^{¢}\left(x\right)=e^x$

Since $M(x,y)=\frac{\partial f}{\partial x}$.

$g\left(x\right)=e^x$

$g^{¢}\left(x\right)=e^x$

Then we have

localid="1668182384287" $g^{¢}\left(x\right)=e^x$

Substitute initial condition $y\left(0\right)=1$.

$\begin{array}{c}2+e-e+1=c\\ c=3\end{array}$

Hence, the required solution is,$2y+xy+y{e}^y-e^y+e^x=3$