Question

In Problems 31-36 solve the given differential equation by finding as in Example 4, an appropriate integrating factor.

$\mathbf{32}\mathbf{.}\mathit{y}\mathbf{(}\mathit{x}\mathbf{+}\mathit{y}\mathbf{+}\mathbf{1}\mathbf{)}\mathit{d}\mathit{x}\mathbf{+}\mathbf{(}\mathit{x}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{)}\mathit{d}\mathit{y}\mathbf{=}\mathbf{0}$

Short answer

The resulting equation is $e^x\left(xy+y^2\right)=c$.

Step-by-step solution

Step 1: Identifying M and N from differential equation  

$\begin{array}{l}M=xy+y^2+y\\ N=x+2y\end{array}$

Finding N_xandN_y

$\begin{array}{l}{M}_y=x+2y+1\\ N_x=1\\ \frac{M_y-N_x}{N}=\frac{x+2y+1-1}{x+2y}=\frac{x+2y}{x+2y}=1\\ \mu \left(x\right)=e^{\int dx}=e^x\end{array}$

Since $\frac{M_y-N_x}{N}$does not depend on more than a variable,

the integrating factor$=e^{\int \frac{M_y-N_x}{N}dx}$

$e^x\left(xy+y^2+y\right)dx+e^x(x+2y)dy=0$

Multiply both sides of the differentia equation by the integrating factor

$\begin{array}{l}M(x,y)=e^x\left(xy+y^2+y\right)\\ \leftrightarrow \frac{\partial M}{\partial y}=e^x(x+2y+1)\\ N(x,y)=e^x(x+2y)\\ \leftrightarrow \frac{\partial N}{\partial x}=e^x(x+2y+1)\end{array}$

Checking whether the differential equation

is exact

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

Therefore, the equation is exact

Find the integrate

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

Integrate

$\begin{array}{l}N(x,y)=\frac{\partial f}{\partial y}=e^x(x+2y)\\ \frac{\partial f}{\partial x}=e^x\left(xy+y^2+y\right)+g^{\text{'}}\left(x\right)\end{array}$

Take

$\begin{array}{l}\frac{\partial f}{\partial x}\\ g\text{'}\left(x\right)=0\\ g\left(x\right)=c\end{array}$

Final proof

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

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

Final solution$e^x\left(xy+y^2\right)=c$