Answer to Question #136653 in Differential Equations for Nikhil

Question

Solve
x(x^2+y^2-a^2)dx+y(x^2+y^2-b^2)dy=0

Answer 1

$\displaystyle x(x^2+y^2-a^2)dx+y(x^2+y^2-b^2)dy =0\\ P = x(x^2+y^2-a^2)dx, Q = y(x^2+y^2-b^2)dy\\ \frac{\partial P}{\partial y} = 2xy = \frac{\partial Q}{\partial x}\\ \textsf{Since}\hspace{0.1cm} \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}, \hspace{0.1cm} \textsf{the ODE is exact}\\ \begin{aligned} z &= \int x(x^2+y^2-a^2) \hspace{0.1cm}\mathrm{d}x \\&= \frac{x^4}{4} + \frac{x^2 y^2}{2} - \frac{x^2 a^2}{2} + f(y) \end{aligned}\\ \begin{aligned} z &= \int y(x^2+y^2-b^2)\hspace{0.1cm}\mathrm{d}y \\& = \frac{y^4}{4} + \frac{x^2 y^2}{2} - \frac{b^2 y^2}{2} + f(x) \end{aligned} \\ \therefore z = \frac{x^4 + y^4}{4} + \frac{x^2 y^2}{2} - \frac{x^2 a^2 + b^2 y^2}{2} \\\textsf{is a solution to the ODE}$