Question

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

$\mathit{x}\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{=}\mathbf{2}\mathit{x}{\mathit{e}}^{\mathbf{x}}\mathbf{-}\mathit{y}\mathbf{+}\mathbf{6}{\mathit{x}}^{\mathbf{2}}$

Short answer

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

Step-by-step solution

Given information

The given differential equation is

$x\frac{dy}{dx}=2x{e}^x-y+6x^2$

Multiply both side by $dx$

$xdy=\left(2x{e}^x-y+6x^2\right)dx$

Move the $dx$term to the left side of the equation

$\left(-6x{}^2+y-2x{e}^x\right)dx+xdy=0$

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

$\begin{array}{l}M(x,y)=-6x{}^2+y-2x{e}^x\\ N\left(x,y\right)=x\end{array}$

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

To take a partial derivative with respect to $y$of $M\left(x,y\right)$

$\begin{array}{l}M(x,y)=-6x{}^2-y-2x{e}^x\\ \frac{\partial M}{\partial y}=1\end{array}$

To take a partial derivative with respect to $x$of $N\left(x,y\right)$

$\begin{array}{l}N\left(x,y\right)=x\\ \frac{\partial N}{\partial x}=1\end{array}$

We can see the condition holds true

$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}=1$

Because $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$there exists a function $f\left(x,y\right)$such that $\frac{\partial f}{\partial x}=M(x,y)$and$\frac{\partial f}{\partial y}=N(x,y)$

$\begin{array}{l}\frac{\partial f}{\partial x}=-6x+y-2x{e}^x\\ \frac{\partial f}{\partial y}=x\end{array}$

Integrate the equations

Take the integral of both sides of the equation for $\frac{\partial f}{\partial x}$with respect to$x$so that we can obtain $f\left(x,y\right)$.

$\int \left(\frac{\partial f}{\partial x}\right)dx=\int \left(-6x^2+y-2x{e}^x\right)dx+\text{g}\left(\text{y}\right)$

The left side is just$f\left(x,y\right)$while the right side we split into three integrals

$f(x,y)=\int -6x^{2}dx+\int ydx-\int 2x{e}^{x}dx+g\left(y\right)$

The two integrals on the left are fairly simple, and we can remove the constant term from the integral on the far right.

$f(x,y)=-2x^3+xy-2\int x{e}^{x}dx+g\left(y\right)$

Choose $u$and$dv$to use integration by parts.

$\begin{array}{c}u=x\\ dv=e^{x}dx\end{array}$

$\begin{array}{c}du=dx\\ v=e^x\end{array}$

The form of integration by parts.

$f(x,y)=-2x^3+xy-2\left[uv-\int vdu\right]+g\left(y\right)$

Substituting in u and v and du.

$f(x,y)=-2x^3+xy-2\left[x{e}^x-\int e^{x}dx\right]+g\left(y\right)$

Which gives us $f(x,y)$

$f(x,y)=-2x^3+xy-2x{e}^x+2e^x+g\left(y\right)$

Now take the partial derivative of f with respect to $y$.

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

And set it equal to $\frac{\partial f}{\partial y}=N(x,y)$from above

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

From this we can see the value of $g^{\text{'}}\left(x\right)$.

$g^{\text{'}}\left(x\right)=0$

Now we need to find $g\left(x\right)$

$\int \left[g^{\text{'}}\left(x\right)\right]dx=\int \left(0\right)dx$

This indicates$g\left(x\right)$is a constant

$g\left(x\right)=C$

Now we can see more specifically $f\left(x,y\right)$

$f(x,y)=-2x^3+xy-2x{e}^x+2e^x+C$

$f(x,y)=-2x^3+xy-2x{e}^x+2e^x+C=C$

Combining the constants on one side, we still have just another arbitrary constant. This is the family of solutions to the differential equation.

$-2x^3+xy-2x{e}^x+2e^x=C$

It is possible to find an explicit solution also.

$\begin{array}{c}xy=2x^3+2x{e}^x-2e^x+C\\ y=2x^2+2e^x-2e^x{x}^{-1}+C{x}^{-1}\end{array}$

The result is

$\begin{array}{l}-2x^3+xy-2x{e}^x+2e^x=C\\ or\\ y=2x^2+2e^x-2e^x{x}^{-1}+C{x}^{-1}\end{array}$