Question

In Problems 1-20 determine whether the given differential equation is exact. If it is exact, solve it.

$\left(3x^{2}y+e^y\right)\mathit{d}\mathit{x}\mathbf{+}\left(x^3+x{e}^y-2y\right)\mathit{d}\mathit{y}\mathbf{=}\mathbf{0}$

Short answer

The solution is$x^{3}y+x{e}^y-y^2=c$

Step-by-step solution

Given information

The given equation is,$\left(3x^{2}y+e^y\right)dx+\left(x^3+x{e}^y-2y\right)dy=0$

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

$$\begin{gathered} M(x,y)=3x^{2}y+e^y \\ N(x,y)=x^3+x{e}^y-2y \end{gathered}$$

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 $\frac{\partial M}{\partial y},\frac{\partial N}{\partial x}$

$\begin{array}{c}\frac{\partial M}{\partial y}=3x^{2}y+e^y\\ \frac{\partial N}{\partial x}=3x^2+e^y\end{array}$

The equation is exact, because $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$

Integrate the equation

Integrate

$M(x,y)=\frac{\partial f}{\partial x}=3x^{2}y+e^y$

$f\left(x,y\right)=x^{3}y+x{e}^y+g\left(y\right)$

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

$\frac{\partial f}{\partial y}=x^{3}y+x{e}^y+g\text{'}\left(y\right)$

Since $N(x,y)=\frac{\partial f}{\partial y},$we see that

$g^{\text{'}}\left(x\right)=-2y$

The solution is$x^{3}y+x{e}^y-y^2=c$