Question

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

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

Short answer

The integrating factor is$x^2{y}^2+x^3=c$

Step-by-step solution

To Find the differential equation and differential exact

The given equation is$M=2y^2+3x$.

Identifying $M$and N from the given equation

$\begin{array}{l}M=2y^2+3x\\ N=2xy\end{array}$

$\begin{array}{c}{M}_y=4y\\ N_x=2y\end{array}$

Compute $\frac{M_y-N_x}{N}$

$\begin{array}{c}\frac{M_y-N_x}{N}=\frac{4y-2y}{2xy}\\ =\frac{1}{x}\end{array}$

Then, the integrating factor is

$\begin{array}{c}\mu \left(x\right)=e^{\frac{1}{x}dx}\\ =e^{\ln x}\\ =x\end{array}$

Since$\frac{M_y-N_x}{N}$ only depends on x . So, the integrating factor is x .

Now, multiplying the given equation by the integrating factor x .

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

Multiply both sides of the differential equation by the integrating factor.

$\begin{array}{l}M(x,y)=2x{y}^2+3x^2\\ N(x,y)=2x^{2}y\end{array}$

Checking whether the obtained differential equation is exact or not.

$\begin{array}{l}\frac{\partial M}{\partial y}=4xy\\ \frac{\partial N}{\partial x}=4xy\end{array}$

Therefore, the equation is exact as both are equal

Find the solution

Then, we have $f(x,y)=x^2{y}^2+g\left(x\right)$.

Now, integrate $N(x,y)=\frac{\partial f}{\partial y}$.

$\begin{array}{c}N(x,y)=\frac{\partial f}{\partial y}\\ =2x^{2}y\end{array}$

$\frac{\partial f}{\partial x}=2x{y}^2+g^{¢}\left(x\right)$

Take $\frac{\partial f}{\partial x}$. Then we have,

As, $g\left(x\right)=x^3$

$g^{¢}\left(x\right)=3x^2$

Hence, the solution of the given differential equation is$x^2{y}^2+x^3=c$