Question

In Problems determine whether the given differential equation is exact.

If it is exact, solve it.

$\left(2x{y}^2-3\right)\mathit{d}\mathit{x}\mathbf{+}\left(2x^{2}y+4\right)\mathit{d}\mathit{y}\mathbf{=}\mathbf{0}$

Short answer

The given differential equation is exact and the solution is$x^2{y}^2-3x+4y=C$

Step-by-step solution

Describe the given Information

The given equation is$\left(2x{y}^2-3\right)dx+\left(2x^{2}y+4\right)dy=0$. An equation of the form $Mdx+Ndy=f\left(\text{x}\right)$is exact only if it satisfieslocalid="1663835269304" $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$

Determining the exactness of the differential equation

Find M and N using given equation

$\begin{array}{l}M(x,y)=2x{y}^2-3\\ N(x,y)=2x^{2}y+4\end{array}$

Find localid="1663835289385" $\frac{\partial M}{\partial y}$and localid="1663835295301" $\frac{\partial N}{\partial x}$

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

Aslocalid="1663835283116" $\frac{\partial \text{M}}{\partial \text{y}}=\frac{\partial \text{N}}{\partial \text{x}}$the equation is exact

Find the solution of differential equation

The solution of the differential equation is a function f(x,y) such that $\frac{\partial f}{\partial x}=\text{M}$and $\frac{\partial f}{\partial \text{y}}=\text{N}$

$\frac{\partial f}{\partial x}=2x{y}^2-3$and $\frac{\partial f}{\partial y}=2x^{2}y+4$After integrating the first of these equations, we get

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

$N\left(x,y)\right)$is obtained by taking the partial derivative of the previous expression with respect to y and setting the result to $N\left(x,y\right)$

$\begin{array}{c}\frac{\partial f}{\partial y}=2x^{2}y+g\text{'}\left(y\right)\\ =2x^{2}y+4\end{array}$

On comparing these two equations we get

$\begin{array}{c}g\text{'}\left(y\right)=4\\ g\left(y\right)=4y\end{array}$

Therefore, function becomes $f(x,y)=x^2{y}^2-3x+4y$, As a result, the implicit solution of the problem is$x^2{y}^2-3x+4y=C$