Question

Determine whether the given differential equation is exact. If it is exact, solve it.

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

Short answer

The solution is$x{y}^3+\frac{x^4}{4}=c$

Step-by-step solution

Given information

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

Determining the exactness of the differential equation

Find M and N using given equation

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

Find $\begin{array}{ll}\frac{\partial M}{\partial y}& \end{array}$and $\begin{array}{l}\frac{\partial N}{\partial x}\\ \end{array}$.

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

As $\frac{\partial \text{M}}{\partial \text{y}}=\frac{\partial \text{N}}{\partial \text{x}}$so the equation is exact

Integrate the equation

The solution of the differential equation is a function$f\left(x,y\right)$such that$\frac{\partial f}{\partial x}=\text{M}$and $\frac{\partial f}{\partial \text{y}}=\text{N}$. Here $\frac{\partial f}{\partial x}=x^3+y^3$and $\frac{\partial f}{\partial y}=3x{y}^2$.After integrating the first of these equations, we get$f(x,y)=\frac{{\text{x}}^{\text{4}}}{\text{4}}\text{+g}\left(\text{y}\right)$

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

$\begin{array}{c}\frac{\partial f}{\partial y}=g^{\text{'}}\left(y\right)\\ =3x^{2}y\end{array}$

On comparing these two equations we get

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

$x{y}^3+\frac{x^4}{4}=c$

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