Answer to Question #129865 in Differential Equations for Ankit

Question

q+xp=p^2

Answer 1

q+xp-p²=0

$\frac{∂f} {∂p}$ =(x-2p)

$\frac{∂f} {∂q}$ =1

$\frac{∂f} {∂x}$ =p

$\frac{∂f} {∂y}$ =0

$\frac{∂f} {∂z}$ =0

$\frac{dx} {(-∂f/∂p)}$ =$\frac{dy} {(-∂f/∂p)}$ =$\frac{dz} {(-p ∂f/∂p-q ∂f/∂q)}$ =

$\frac {dp} {(∂f/∂x+p ∂f/∂z)}$ +$\frac{dq} {(∂f/∂y+q ∂f/∂z)}$ =$\frac{(∂∅)} {0}$

$\frac{dx} {(-(x-2p) )}$ =

$\frac{dy} {(-1)} =\frac{dz} {(-p(x-2p)-q)} =\frac{dp} {p} =\frac{dq} {0} =\frac{(∂∅)} {0}$

=$\frac {dy} {(-1)} =\frac{dp} {p}$

log ⁡p=-y+log ⁡a

p=ae^-y

q+xae^-y=(ae^-y)²

q=a² e^-2y-xae^-y

dz=pdx+qdy

dz=ae^-y dx+a² e^-2y-xae^-y dy

dz=a e^-y dx-xe^-y dy)+a² e^-2y dy

Using −p(x−2p)−q=p²=a²e^2y the z and p fractions combine to dz=pdp which integrates to

$z=axe^{-y}-\frac{1}{2}a^2e^{-2y}+b$