Answer to Question #145388 in Differential Equations for Jisha

Question

Solve : (x^3+xy^2+a^2y) dx+(y^3+yx^2+a^2x) dy=0

Answer 1

$\displaystyle (x^3+xy^2+a^2y)\mathrm{d}x + (y^3+yx^2+a^2x)\mathrm{d}y=0\\ x^3\mathrm{d}x + xy^2\mathrm{d}x + a^2y\mathrm{d}x + y^3\mathrm{d}y + yx^2\mathrm{d}y + a^2x \mathrm{d}y =0\\ x^3 \mathrm{d}x + y^3 \mathrm{d}y + xy^2\mathrm{d}x + yx^2\mathrm{d}y + a^2(y\mathrm{d}x + x\mathrm{d}y)=0\\ x^3 \mathrm{d}x + y^3 \mathrm{d}y + xy(y\mathrm{d}x + x\mathrm{d}y) + a^2(y\mathrm{d}x + x\mathrm{d}y)=0\\ x^3 \mathrm{d}x + y^3 \mathrm{d}y + (xy + a^2)(y\mathrm{d}x + x\mathrm{d}y) =0\\ \textsf{Integrating both sides}\\ \int x^3\, \mathrm{d}x + \int y^3 \,\mathrm{d}y + \int\,(xy + a^2)\mathrm{d}(xy) = C\\ \textsf{Substitute}\, u = xy\,\textsf{in the last integral}.\\ \int x^3\, \mathrm{d}x + \int y^3 \,\mathrm{d}y + \int\,(u + a^2)\mathrm{d}u = C\\ \frac{x^4}{4} + \frac{y^4}{4} + \frac{u^2}{2} + a^2u = C\\ \frac{x^4}{4} + \frac{y^4}{4} + \frac{(xy)^2}{2} + a^2xy = C\\ x^4 + y^4 + 2(xy)^2 + a^2xy = C\\ \therefore(x^2 + y^2)^2 + a^2xy = C \,\,\textsf{is a solution to the ODE}\\ \textsf{Where}\, C\, \textsf{is an arbitrary constant.}$