Question

In Problems determine whether the given differential equation is exact. If it is exact, solve it.

$\mathbf{(}\mathbf{5}\mathit{x}\mathbf{+}\mathbf{4}\mathit{y}\mathbf{)}\mathit{d}\mathit{x}\mathbf{+}\mathbf{(}\mathbf{4}\mathit{x}\mathbf{-}\mathbf{8}{\mathit{y}}^{\mathbf{3}}\mathbf{)}\mathit{d}\mathit{y}\mathbf{=}\mathbf{0}$

Short answer

The given differential equation is exact and the solution is$\frac{5}{2}{x}^2+4xy-2y^4=C$

Step-by-step solution

Define exact differential equation

The given equation is$(5x+4y)dx+(4x-8y^3)dy=0$ .An equation of the form $Mdx+Ndy=f\left(\text{x}\right)$is exact only if it satisfieslocalid="1663835002979" $\frac{\partial \text{M}}{\partial \text{y}}=\frac{\partial \text{N}}{\partial \text{x}}$

Determining the exactness of the differential equation

Find M and N using given equation

$\begin{array}{l}M(x,y)=5x+4y\\ N(x,y)=4x-8y^3\end{array}$

Find localid="1663835015863" $\frac{\partial \text{M}}{\partial \text{y}}$and localid="1663835022289" $\frac{\partial \text{N}}{\partial \text{x}}$

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

As localid="1663835035546" $\frac{\partial \text{M}}{\partial \text{y}}=\frac{\partial \text{N}}{\partial \text{x}}$, so, 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 $\frac{\partial f}{\partial \text{y}}=\text{N}$. Here $\frac{\partial f}{\partial x}=5x+4y$ and $\frac{\partial f}{\partial y}=4x-8y^3$. After integrating the first of these equations, we get$f(x,y)=\frac{5}{2}{x}^2+4xy+g\left(y\right)$.

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

$\begin{array}{c}\frac{\partial f}{\partial y}=4x+g\text{'}\left(y\right)\\ =4x-8y^3\end{array}$

On comparing these two equations we get

$\begin{array}{c}g\text{'}\left(y\right)=4x-8y^3\\ g\left(y\right)=-2y^4\end{array}$

Therefore, function $f(x,y)$ becomes $f(x,y)=\frac{5}{2}{x}^2+4xy-2y^4$. As a result, the implicit solution of the problem is $\frac{5}{2}{x}^2+4xy-2y^4=C$