Answer to Question #255546 in Differential Equations for Sourav

Question

Find the complete integral of the differential equation

p^2- y³q = x² -y².

Answer 1

$\frac{dx}{-f_p}=\frac{dy}{-f_q}=\frac{dz}{-pf_p-qf_q}=\frac{dp}{f_x+pf_z}=\frac{dq}{f_y+qf_z}$

$\frac{dx}{-2p}=\frac{dy}{y^3}=\frac{dz}{2p^2-y^3q}=\frac{dp}{-2x}=\frac{dq}{2y}$

$2xdx=2pdp$

$p^2=x^2+c$

$y=c_1$

$2ydy/y^3=dq$

$q=-2/y+c'$

$\frac{d(\sqrt{x^2+c})}{-2x}=\frac{d(-2/y+c')}{2y}$

case 1:

$-\frac{dx}{2\sqrt{x^2+c}}=\frac{dy}{y^3}$

$-1/2y^2=-ln(\sqrt{x^2+c}+x)/2+c'_2$

$c_2=y^2-ln(\sqrt{x^2+c}+x)$

$F(c_1,c_2)=F(y,y^2-ln(\sqrt{x^2+c}+x))=0$

case 2:

$\frac{-d(\sqrt{x^2+c})}{-2x}=\frac{d(-2/y+c')}{2y}$

$-1/2y^2=ln(\sqrt{x^2+c}+x)/2+c'_2$

$c_2=y^2+ln(\sqrt{x^2+c}+x)$

$F(c_1,c_2)=F(y,y^2+ln(\sqrt{x^2+c}+x))=0$