x(x2+y2−a2)dx+y(x2+y2−b2)dy=0P=x(x2+y2−a2)dx,Q=y(x2+y2−b2)dy∂P∂y=2xy=∂Q∂xSince∂P∂y=∂Q∂x,the ODE is exactz=∫x(x2+y2−a2)dx=x44+x2y22−x2a22+f(y)z=∫y(x2+y2−b2)dy=y44+x2y22−b2y22+f(x)∴z=x4+y44+x2y22−x2a2+b2y22is a solution to the ODE\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}x(x2+y2−a2)dx+y(x2+y2−b2)dy=0P=x(x2+y2−a2)dx,Q=y(x2+y2−b2)dy∂y∂P=2xy=∂x∂QSince∂y∂P=∂x∂Q,the ODE is exactz=∫x(x2+y2−a2)dx=4x4+2x2y2−2x2a2+f(y)z=∫y(x2+y2−b2)dy=4y4+2x2y2−2b2y2+f(x)∴z=4x4+y4+2x2y2−2x2a2+b2y2is a solution to the ODE
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