3.3.19.1 Example

$$\begin{array}{rl}ydx+x(x^{2}y-1)dy& =0\\ M(x,y)+N(x,y)y^{\prime }& =0\end{array}$$

Where

$$\begin{array}{rl}\frac{\partial M}{\partial y}& =1\\ \frac{\partial N}{\partial x}& =3x^{2}y-1\end{array}$$

Hence not exact. Trying the above 3 methods shows it is not possible to find an integrating factor. But by inspection let $I=\frac{y}{x^3}$. Then the ode becomes

$$\begin{array}{rl}yIdx+Ix(x^{2}y-1)dy& =0\\ y\frac{y}{x^3}dx+\frac{y}{x^3}x(x^{2}y-1)dy& =0\\ \frac{y^2}{x^3}dx+(y^2-\frac{y}{x^2})dy& =0\\ M(x,y)+N(x,y)y^{\prime }& =0\end{array}$$

Where

$$\begin{array}{rl}M& =\frac{y^2}{x^3}\\ N& =(y^2-\frac{y}{x^2})\end{array}$$

Now we see that the ode is exact by checking:

$$\begin{array}{rl}\frac{\partial M}{\partial y}& =\frac{2y}{x^3}\\ \frac{\partial N}{\partial x}& =-(-2\frac{y}{x^3})=\frac{2y}{x^3}\end{array}$$

Since ode is now exact, we need to find $\varphi$ from

$$\begin{array}{}\text{(3)}& \frac{\partial \varphi }{\partial x}& =M\text{(4)}& \frac{\partial \varphi }{\partial y}& =N\end{array}$$

From (3)

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

Therefore

$$\begin{array}{rl}\varphi & =\int Mdx+f\left(y\right)\\ & =\int \frac{y^2}{x^3}dx+f\left(y\right)\\ & =y^2\int x^{-3}dx+f\left(y\right)\\ & =y^2\frac{x^{-2}}{-2}+f\left(y\right)\\ \text{(5)}& & =\frac{y^2}{-2x^2}+f\left(y\right)\end{array}$$

Where $f\left(y\right)$ is arbitrary function to be found. Taking derivative of the above w.r.t. $y$ gives

$$\begin{array}{rl}\frac{\partial \varphi }{\partial y}& =\frac{d}{dy}(-\frac{y^2}{2x^2}+f\left(y\right))\\ & =-\frac{y}{x^2}+f^{\prime }\left(y\right)\end{array}$$

Comparing the above to (4) shows that

$$\begin{array}{rl}N& =-\frac{y}{x^2}+f^{\prime }\left(y\right)\\ y^2-\frac{y}{x^2}& =-\frac{y}{x^2}+f^{\prime }\left(y\right)\\ f^{\prime }\left(y\right)& =y^2\end{array}$$

Hence

$$\begin{array}{rl}f\left(y\right)& =\int y^{2}dy\\ & =\frac{y^3}{3}+c\end{array}$$

Substituting this into (5) gives

$$\begin{array}{rl}\varphi & =\frac{y^2}{-2x^2}+f\left(y\right)\\ & =\frac{y^2}{-2x^2}+\frac{y^3}{3}+c\end{array}$$

Since $\varphi$ is also constant function then we can simplify the above to

$$\begin{array}{rl}\frac{y^2}{-2x^2}+\frac{y^3}{3}& =C\\ 3y^2-2x^2{y}^3& =6x^{2}C\\ 3y^2-2x^2{y}^3& =x^2{C}_1\end{array}$$