How to find p and q (Charpit method) for this equation

[lotusquantum]

Homework Statement

$$2uq^2+y^2(q-p)=0$$
$$p= \partial u/\partial x$$; $$q= \partial u/\partial y$$
find the general solution

Homework Equations

We get the charpit equations:
$$\frac{dx}{-y^2}=\frac{dy}{4uq+y^2}=\frac{du}{-py^2+q(4uq+y^2)}=\frac{dp}{-2pq^2}=\frac{dq}{-2y(q-p)-2q^3}$$

I don't know how to find the value of p and q ??

The Attempt at a Solution

(i have tried the following ways but i cannot:
from the original equation we have: $$(q-p) =-\frac{2uq^2}{y^2}$$
substitute into the last pair equation $$\frac{dq}{-2y(-\frac{2uq^2}{y^2})-2q^3}=\frac{dp}{-2pq^2}$$
or $$\frac{dq}{dp}=\frac{1}{p}(\frac{-2u}{y}+q) => q = \frac{2u}{y}+p.c1$$
but i don't know how to find p? because any pairs will lead to a complicated equation. please experts help! thanks )

 

[mighty2000]

To find the values of p and q, we can use the method of characteristics. This involves solving the system of equations given by the charpit equations.

First, we can rearrange the original equation to get:
2uq^2 + y^2q - y^2p = 0

Now, using the charpit equations, we can set up the following system of equations:
dx = -y^2 ds
dy = 4uq + y^2 ds
du = -py^2 + q(4uq + y^2) ds
dp = -2pq^2 ds
dq = -2y(q-p) - 2q^3 ds

We can then solve for dp/ds and dq/ds by substituting the values of dx and dy into the equations for dp and dq. This gives us:
dp/ds = 2y^2p + 8uq^2 + 4q^3
dq/ds = 2y(q-p) - 4q^3

Next, we can substitute these values into the equations for du and dq to get:
du/ds = -py^2 + q(4uq + y^2)
dq/ds = -2y(q-p) - 2q^3

Now, we can solve for p and q by setting the equations for du/ds and dq/ds equal to 0. This gives us:
0 = -py^2 + q(4uq + y^2)
0 = -2y(q-p) - 2q^3

Solving for p and q in terms of u and y, we get:
p = \frac{q(4uq + y^2)}{y^2}
q = -\frac{2y(q-p)}{2q^2 + y^2}

Substituting these values back into the original equation, we get:
2uq^2 + y^2q - y^2p = 0
2uq^2 + y^2q - y^2(\frac{q(4uq + y^2)}{y^2}) = 0
2uq^2 + y^2q - 4uq^2 - y^2q = 0
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