Answer to Question #254308 in Differential Equations for Khalesy

Question

yp-x²q²-x²y=0

Answer 1

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

$f_p=y,f_q=-2qx^2,f_x=-2xq^2-2xy,f_y=p-x^2,f_z=0$

$\frac{-2x^2dx+qdy+xdp}{2x^2y+2q^2x^2-2x^2y-2q^2x^2}=\frac{-2x^2dx+qdy+xdp}{0}$

$-2x^2dx+qdy+xdp=0$

$(p-x^2)dx=-ydq$

$-3x^2dx+ydq+qdy+xdp+pdx=0$

$-3x^2dx+d(qy)+d(px)=0$

$-x^3+qy+px=c$

$\frac{dx}{x}=\frac{dy}{y}=\frac{dz}{x^3+c}$

$lny=lnx+lnc_1$

$c_1=y/x$

$z=x^3/3+clnx+c_2$

$F(c_1,c_2)=F(y/x,z-x^3/3-clnx)=0$