Use the methods of this section to solve the following differential equations. Compare computer solutions and reconcile differences.

$\mathit{x}\mathit{y}\mathbf{\text{'}}\mathbf{}\mathbf{+}\mathbf{}\mathit{y}\mathbf{}\mathbf{=}\mathbf{}{\mathit{e}}^{\mathbf{x}\mathbf{y}}\mathbf{}\mathit{H}\mathit{i}\mathit{n}\mathit{t}\mathbf{:}\mathbf{}\mathit{L}\mathit{e}\mathit{t}\mathbf{}\mathit{u}\mathbf{=}\mathit{x}\mathit{y}$

The given differential equation is $xy\text{'}+y=e^{xy}$.

A differential equation is a mathematical equation for an unknown function of one or several variables that relates the values of the function itself to its derivatives of various orders.

Let u = xy, then u' = y + xy', so the differential equation becomes

$$\begin{gathered} u\text{'}-y+y=e^u \\ du-e^{u}dx=0 \end{gathered}$$

we can show that this is not an exact equation,

$$\begin{gathered} \frac{\partial }{\partial u}\left(-e^u\right)=-e^u\ne \frac{\partial }{\partial u}\left(1\right)=0 \\ \mu du-\mu e^{u}dx=0 \end{gathered}$$

And because this integrating factor will make it exact

$\begin{array}{rcl}\frac{\partial }{\partial u}\left(-\mu e^u\right)& =& \frac{\partial }{\partial x}\left(\mu \right)-\mu \text{'}{e}^u-\mu e^u=0\\ \frac{d\mu }{\mu }& =& -du\\ \mu & =& e^{-u}\end{array}$

So, the differential equation becomes an exact equation if we multiply it by e^-u,

$\begin{array}{rcl}{e}^{-u}du-dx& =& 0\end{array}$

And we can check

$\begin{array}{rcl}\frac{\partial }{\partial u}\left(-1\right)& =& 0=\frac{\partial }{\partial x}\left(e^{-u}\right)\end{array}$

a function F (x,u) such that

$\begin{array}{rcl}-1& =& \frac{\partial F}{\partial x},e^{-u}=\frac{\partial F}{\partial u},-dx+e^{-u}du=dF\end{array}$

Take $\begin{array}{rcl}-1& =& \partial F/\partial x\end{array}$and solve it (n.b. we can take$\begin{array}{rcl}{e}^{-u}& =& \partial F\backslash \partial u\end{array}$instead)

$\begin{array}{rcl}\int dF& =& -\int dxF=-x+f\left(y\right)\end{array}$

Function f(u) we are going to integrate F(x,u) with respect to u,

$\begin{array}{rcl}\frac{\partial F}{\partial u}& =& f\text{'}\left(u\right)\end{array}$

But, we know that $\begin{array}{rcl}\partial F/\partial u& =& e^{-u}\end{array}$, so

$\begin{array}{rcl}{e}^{-u}& =& f\text{'}\left(u\right)\\ f\left(u\right)& =& \int e^{-u}du=-e^{-u}+C\end{array}$

Therefore, the function F (x,u) is

$\begin{array}{rcl}F\left(x,u\right)& =& -x-e^{-u}+C\end{array}$

then $\begin{array}{rcl}-dx+e^{-u}du& =& dF=0\end{array}$, and therefore, for the equation $\begin{array}{rcl}-dx+e^{-u}du& =& 0\end{array}$the solution is

$\begin{array}{rcl}F\left(x,u\right)& =& C_1\\ \therefore x+e^{-u}& =& C_2\end{array}$

Finally, substitute u = xy back into the solution, so we get

$\begin{array}{rcl}x+e^{-xy}& =& C_2,y=-\frac{In\left(C_2-x\right)}{x}\end{array}$