Answer to Question #139604 in Differential Equations for Lavanya

Question

X^2p^2+xyp+2y^2=0

Answer 1

Given

$$x^2p^2+xyp+2y^2=0$$

Let us denote $xp=q$ ,thus we get $q^2+yq+2y^2=0$ , now solving q ,we get,

$$q=\frac{-y\pm\sqrt{y^2-4\cdot 1\cdot 2y^2}}{2}\\ \implies q=\frac{-y\pm y\sqrt{7}i}{2}\\ \implies xp=x\frac{dy}{dx}=y(-1\pm\sqrt{7}i)\\ \implies \frac{dy}{y}=(-1\pm\sqrt{7}i)\frac{dx}{x}\\ \implies \int\frac{dy}{y}=(-1\pm\sqrt{7}i)\int\frac{dx}{x}\\ \implies \ln y=-1\pm\sqrt{7}i\ln x+\ln c\\ \implies y=cx^{-1\pm\sqrt{7}i}$$

Where c is constant and $i=\sqrt{-1}$