I have been solving a pde for real $a$,$b$.
$pq^2=ax+by$, where $p=\frac{dz}{dx}$, $q=\frac{dz}{dy}$.
I am required to find the first integral of the equation: $F(z,x,y,c_1,c_2)=0$
I wrote down the Charpit's/auxiliary equations:
$\frac{dp}{a}=\frac{dq}{b}=\frac{dz}{3q^2p}=\frac{dx}{q^2}=\frac{dy}{2qp}$.
In the course of solving which I have arrived at the following system of equations (1):
$dz=pdx+qdy$
$z=\frac{3b^2p^4}{4a^3}-\frac{2bc_1p^3}{a^3}+\frac{3c_1^2p^2}{2a^3}+c_5$;
$x=\frac{b^2p^3}{3a^3}-\frac{bc_1p^2}{a^3}+\frac{(c_1)^2p}{a^3}+c_4$;
$y=\frac{2bp^3}{3a^2}-\frac{c_1p^2}{a^2}-c_3$,
where $c_1,c_3,c_4$ and $c_5$ are constants which should add up to two constants $c_1$ and $c_2$ in the end. As you can see, this system of equations is only not resolved with respect to $p$, which is the only term I haven't been able to write in the form of $g(z,x,y)$.
The system (1) is correct when substituted into the equation, I've checked. The only fraction i haven't used in any way is the third one: $\frac{dz}{3q^2p}=\frac{dz}{3(\frac{bp-c_1}{a})^2p}$.
The equation for $x$ is a non-linear first order ODE, because there is no $y$ in the equation, but Wolfram Alpha does not solve it, for whatever reason. If it can be solved with respect to $z$ ($p=\frac{dz}{dx}$), then my problem is solved. How could i do that? If it can be done, then my immediate follow-up question is the following: why didn't i need to use the fraction $\frac{dz}{3q^2p}=\frac{dz}{3(\frac{bp-c_1}{a})^2p}$ in the course of solving the problem?
I had also obtained the following equations:
$bp=aq+c_1$ or $q=\frac{bp-c_1}{a}$,
$dy=dp(\frac{2bp^2}{a^2}-\frac{2pc_1}{a^2})$ and
$dx=\frac{dp}{a}(\frac{bp-c_1}{a})^2$.
I've tried to somehow arrive at $z=f(x,y,c_1,c_2)$ or $F(z,x,y,c_1,c_2)=0$, but had no success because the system (1) is not nicely linear and terms with $p$ always crawl into calculations. I'm now out of ideas.
Could anybody perhaps help me in some way? Or help me solve the ODE $x=\frac{b^2p^3}{3a^3}-\frac{bc_1p^2}{a^3}+\frac{(c_1)^2p}{a^3}+c_4$? Thank you!
Best Answer
@Nick The Dick. Your calculus is correct.
An equation not used is : $$\frac{dq}{b}=\frac{dz}{3q^2p}=\frac{dz}{3q^2(\frac{aq+c_1}{b})}$$ $$dz=3b^2q^2(aq+c_1)dq$$ $$z=\frac34 ab^2q^4+c_1b^2q^3+C$$ $$z=\frac34 ab^2q^4+(bp-aq)b^2q^3+C$$ $$z=b^2q^3(bp-\frac14 aq)+C$$