Solving auxiliary equations in Charpit's method. Non-linear first-order ODE.

Question

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!

Answer 1

First of all I am introducing another method to solve the given pde. After that I will explain your query (Although I could edit the answer I gave earlier, yet I think it is important to keep it in order to accommodate your query).

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Given pde is$$pq^2=ax+by\qquad\text{where}~~p=\frac{\partial z}{\partial x},~q=\frac{\partial z}{\partial y}\tag1$$ Let $~f(x,y,z,p,q)\equiv pq^2-ax-by=0~$ and $~X=ax+by~$.

Now $$p=\frac{\partial z}{\partial x}=\frac{\partial z}{\partial X}\cdot\frac{\partial X}{\partial x}=aP~~~\qquad~~~~~~~~~~~~~~~~~~~~~~~~~$$ $$q=\frac{\partial z}{\partial y}=\frac{\partial z}{\partial X}\cdot\frac{\partial X}{\partial y}=bP~,\qquad\text{where}~~P=\frac{\partial z}{\partial X}$$ From $(1)$, $$pq^2=ax+by=X$$ $$\implies aP\cdot b^2P^2=X$$ $$\implies P=\left(\dfrac{X}{ab^2}\right)^{\frac 13}$$ Then $$p=a\cdot \left(\dfrac{X}{ab^2}\right)^{\frac 13}\qquad\text{and}\qquad q=b\cdot\left(\dfrac{X}{ab^2}\right)^{\frac 13}$$ Now $$dz=p~dx+q~dy$$ $$\implies dz=a\cdot \left(\dfrac{X}{ab^2}\right)^{\frac 13}~dx~+~b\cdot\left(\dfrac{X}{ab^2}\right)^{\frac 13}~dy$$ $$\implies dz=\left(\dfrac{X}{ab^2}\right)^{\frac 13}~(a~dx~+~b~dy)$$ $$\implies dz=\left(\dfrac{X}{ab^2}\right)^{\frac 13}~d(ax~+~by)$$ $$\implies dz=\left(\dfrac{X}{ab^2}\right)^{\frac 13}~dX$$ Integrating, $$z=\dfrac 34\dfrac{X^{\frac 43}}{(ab^2)^{\frac 13}}~+c$$ $$\implies z=\dfrac 34\dfrac{1}{a^{\frac13}b^{\frac23}}~(ax+by)^{\frac 43}+c$$where $$ is integrating constant.

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Explanation of your query:

It is clear that the solution of the given pde in both the two cases are same. Although in the first method there should be an arbitrary constant in equation $(2)$ as it is the general rule for integration. Let's see what happened when we take an arbitrary constant in equation $(2)$.

$$~\dfrac{dp}{a}=\dfrac{dq}{b}\implies p=\dfrac ab q+c~\tag{2a}$$where $$ is integrating constant.

From $(1)$ , we have $$\left(\dfrac ab q+c\right)q^2=ax+by$$ $$\implies \dfrac ab q^3+cq^2=ax+by$$ $$\implies q^3+\dfrac {bc}{a} q^2-\dfrac ba\cdot(ax+by)=0$$which is a general cubic equation of $q~$ and can be solved by using Cardano's formula by reduction to a depressed cubic. Here the solution is a laborious task.

Answer 2

Here $~f(x,y,z,p,q)\equiv pq^2-ax-by=0~\tag1$

By Charpit's Method, the auxiliary equations are

$$\dfrac{dx}{f_p}=\dfrac{dy}{f_q}=\dfrac{dz}{pf_p+qf_q}=-\dfrac{dp}{f_x + pf_z}=-\dfrac{dq}{f_y + qf_z}$$ $$\implies\dfrac{dx}{q^2}=\dfrac{dy}{2pq}=\dfrac{dz}{3pq^2}=-\dfrac{dp}{-a}=-\dfrac{dq}{-b}$$

From the last two ratios, $~\dfrac{dp}{a}=\dfrac{dq}{b}\implies p=\dfrac ab q~\tag2$

Putting the value of$~p~$ in $(1)$, we have $$\dfrac ab q^3-ax-by=0\implies q^3=\dfrac ba (ax+by)\implies q= \sqrt[3]{\dfrac{b}{a}~(ax+by)}$$

So $~p=~\dfrac ab~\sqrt[3]{\dfrac{b}{a}~(ax+by)}~.$

Now $$dz=p~dx+q~dy$$ $$\implies dz=\dfrac ab~\sqrt[3]{\dfrac{b}{a}~(ax+by)}~dx~+~\sqrt[3]{\dfrac{b}{a}~(ax+by)}~dy$$ $$\implies dz=~\sqrt[3]{\dfrac{b}{a}~(ax+by)}~\left(\dfrac ab~dx~+~dy\right)$$ $$\implies dz=\dfrac 1b~\sqrt[3]{\dfrac{b}{a}~(ax+by)}~\left(a~dx~+~b~dy\right)$$ $$\implies dz=\dfrac{1}{a^{\frac13}b^{\frac23}}~(ax+by)^{\frac 13}~d(ax~+~by)$$ Integrating we have, $$z=\dfrac 34\dfrac{1}{a^{\frac13}b^{\frac23}}~(ax+by)^{\frac 43}+c$$where $~c~$ is integrating constant.

Answer 3

@Nick The Dick. Your calculus is correct.

An equation not used is : $$\frac{dq}{b}=\frac{dz}{3q^2p}=\frac{dz}{3q^2(\frac{aq+c_1}{b})}$$ $$dz=3b^2q^2(aq+c_1)dq$$ $$z=\frac34 ab^2q^4+c_1b^2q^3+C$$ $$z=\frac34 ab^2q^4+(bp-aq)b^2q^3+C$$ $$z=b^2q^3(bp-\frac14 aq)+C$$