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

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!

• Your first equation should be $dz=pdx+qdy$ not $z=pdx+qdy$ Jan 6, 2020 at 1:59
• @Isham oops sorry, yeah, you're right. But that doesnt help unfortunately :( thank you though:) Jan 6, 2020 at 2:57
• @lsham well it is definitely problematic, especially to me:) but unfortunately I need it in order to get a pass. The professor is gonna probably help me anyway, but I'm pretty upset with our program. The professor doesn't admit some book methods, or either says they're wrong. And about the charpit method, it's really strange how every thing there is about it on YouTube is being explained in a heavy Indian accent. I'm yet to figure out the reason why. I've read somewhere that it's synonymous with the characteristics method, but what I know is that the latter is for linear pdes. Oh well Jan 6, 2020 at 4:03
• @lsham and the reason I said all that is because I'm yet to read somewhere comprehensively (to me) what the method is. Because right now it's "write those equations(the derivation I read somewhere - that's okay) and then somehow magically solve them" :) Jan 6, 2020 at 4:10
• I understand what you mean. Maybe someone will come with an answer. I hope so. Nick. ( I upvote your question for more attention.) Jan 6, 2020 at 4:30

@Nick The Dick. Your calculus is correct.

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

• I haven't managed to solve the problem by manipulating my old system that I wrote in terms of $p$ and $q$ instead of $p$ and $c_1$ and using in addition that new equation for $z$ you've helped me with in your answer. But i can see you've obtained the answer yourself, in the reply to the first answer. Did you do it by using your own calculations not connected to your answer? If not, could you perhaps hint me some more? The other answer doesn't integrate using a constant, and it seems to me to be a mistake. Jan 9, 2020 at 11:50
• I didn't try to solve for the general solution, but only for a particular solution. See the answer from nmasanta in which there was a typo in the three last equations, now corrected if my proposed correction is accepted. Jan 9, 2020 at 14:59

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 $$\dfrac ab q^3-ax-by=0\implies q^3=\dfrac ba (ax+by)\implies q= \sqrt{\dfrac{b}{a}~(ax+by)}$$

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

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

• How do you justify that you don't need a constant of integration during the first integratiion $\frac{dp}{a}=\frac{dq}{b}⟹p=\frac{a}{b}q$? Jan 6, 2020 at 12:30
• @nmasanta. I don't agree with $z=\dfrac 34\left(\dfrac ba\right)^{\frac 23}~(ax+by)^{\frac 43}+c$. I found : $$z=\frac34 a^{-1/3}b^{-2/3}(ax+by)^{4/3}+c$$ Jan 6, 2020 at 13:50
• Can you please show your findings ? @JJacquelin Jan 6, 2020 at 16:54
• Just put the result $z(x,y)$ into $pq^2=ax+by$ and check. Jan 6, 2020 at 18:27
• @NickTheDick I shall justify your query as early as possible. Give me a little more time, actually I am busy in a seminar. Sorry for the inconvenience. Jan 9, 2020 at 16:11

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 $$𝑐$$ is integrating constant.

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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 $$𝑐$$ 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.
• Now define also $Y=-bx+ay$ to have a full coordinate transform and justify why $\frac{∂z}{∂Y}=0$ as you implicitly assume. Jan 14, 2020 at 9:00