Question

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

$\mathbf{33}\mathbf{.}\mathbf{6}\mathbf{xydx}\mathbf{+}\left(4y+9x^2\right)\mathbf{dy}\mathbf{=}\mathbf{0}$

Short answer

The resulting equation is$y^4+3x^2{y}^3=c$

Step-by-step solution

Step 1: Identifying M and N from differential equation  

$\begin{array}{l}N=4y+9x^2\\ M=6xy\end{array}$

Finding N_xandN_y

$\begin{array}{l}{M}_y=6x\\ N_x=18x\\ \frac{M_y-N_x}{N}=\frac{6x-18x}{4y+9x^2}=\frac{-12x}{4y+9x^2}\\ \frac{N_x-M_y}{M}=\frac{18x-6x}{6xy}=\frac{12x}{6xy}=\frac{2}{y}\\ \mu \left(y\right)=e^{\int \frac{2}{y}dy}=e^{2lny}=e^{ln{y}^2}=y^2\end{array}$

Since $\frac{N_x-M_y}{M}$only depends on ,the integrating factor $=e^{\int \frac{N_x-M_y}{M}}dy$

$6x{y}^{3}dx+\left(4y^3+9x^2{y}^2\right)dy=0$

Multiply both sides of the differentia equation by the integrating factor

$\begin{array}{l}M(x,y)=6x{y}^3\\ \leftrightarrow \frac{\partial M}{\partial y}=18x{y}^2\\ N(x,y)=4y^3+9x^2{y}^2\\ \leftrightarrow \frac{\partial N}{\partial x}=18x{y}^2\end{array}$

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

$f(x,y)=y^4+3x^2{y}^3+g\left(x\right)$

Integrate

$\begin{array}{l}N(x,y)=\frac{\partial f}{\partial y}=4y^3+9x^2{y}^2\\ \frac{\partial f}{\partial x}=6x{y}^3+g\text{'}\left(x\right)\end{array}$

Take

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

Final proof

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

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

Final solution

$y^4+3x^2{y}^3=c$