Answer to Question #200708 in Differential Equations for Chavan Kiran

Question

q-px-p^2=0

Answer 1

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

$\frac{dx}{x+2p}=\frac{dy}{-1}=\frac{dz}{px+2p^2-q}=\frac{dp}{-p}=\frac{dq}{0}$

$lnp=y+c$

$p=c_1e^y$

then:

$q=px+p^2=c_1xe^y+c_1^2e^{2y}$

$dz=pdx+qdy=c_1e^ydx+(c_1xe^y+c_1^2e^{2y})dy$

$z=c_1xe^y+c_1xe^y+c_1^2e^{2y}/2+c_2=2c_1xe^y+c_1^2e^{2y}/2+c_2$