Question

In Problems 1-20 determine whether the given differential equation is exact. If it is exact, solve it.

$\left(x^2{y}^3-\frac{1}{1+9x^2}\right)\frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{y}}\mathbf{+}{\mathit{x}}^{\mathbf{3}}{\mathit{y}}^{\mathbf{2}}\mathbf{=}\mathbf{0}$

Short answer

The solution is $\frac{1}{3}{x}^3{y}^3-\frac{1}{3}ta{n}^{-1}\left(3x\right)=C$

Step-by-step solution

Given information

The given equation is$\left(x^2{y}^3-\frac{1}{1+9x^2}\right)\frac{dx}{dy}+x^3{y}^2=0$.

Multiply both side by$dy$.

$\left(x^2{y}^3-\frac{1}{1+9x^2}\right)dx+x^3{y}^{2}dy=0$

Compare the given equation with $Mdx+Ndy=0$

$\begin{array}{c}M(x,y)=x^2{y}^3-\frac{1}{1+9x^2}\\ N(x,y)=x^3{y}^2\end{array}$

Condition for exactness 

An equation of the form $Mdx+Ndy=0$, is said to be exact, if$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$

Determining the exactness of the differential equation

Checking whether the differential equation is exact by finding $\frac{\partial M}{\partial y},\frac{\partial N}{\partial x}$.

$\frac{\partial M}{\partial y}=3x^2{y}^2$

$\frac{\partial N}{\partial x}=3x^2{y}^2$

The differential equation is exact, because $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$.

Next, we have to find the solution of the given equation. Where $f\left(x,y\right)=\frac{1}{3}{x}^3{y}^3+g\left(x\right)$

Find the solution

Integrate $N(x,y)=\frac{\partial f}{\partial y}=x^3{y}^2$.

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

$\begin{array}{l}g\text{'}\left(x\right)=-\frac{1}{1+9x^2}\\ g\left(x\right)=-\frac{1}{3}ta{n}^{-1}\left(3x\right)\\ M(x,y)=\frac{\partial f}{\partial x}\end{array}$

$g\text{'}\left(x\right)=-\frac{1}{1+9x^2}$

Use trigonometric substitution method,

$\begin{array}{c}g\left(x\right)=\frac{1}{3}\cos \theta \\ \frac{1}{3}{x}^3{y}^3-\frac{1}{3}ta{n}^{-1}\left(3x\right)=c\end{array}$

The solution is

$\frac{1}{3}{x}^3{y}^3-\frac{1}{3}ta{n}^{-1}\left(3x\right)=c$