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I know I'm not supposed to ask specific questions so I'll try to generalize it a bit.

I'm studying Calculus in Several Variables and we have not been taught how to solve PDEs nor does it seem like we will be taught any time soon. I've been scouring the internet trying to find how one solves this kind of problem to no avail. Watched tons of videos and read tons of documents. Perhaps I'm stupid, I don't mind admitting that. Here is the problem in question.

$$\frac{\partial}{\partial x}f(x,y)-3x \left( \frac{\partial}{\partial y}f(x,y)\right)=y$$ With the initial condition: $f(x,0)=x^3+x^2$

And a recommendation to change the variables like such: $\begin{cases}u=ax^2+y \\ v=x\end{cases}$

The only thing we have done regarding PDEs within this course has been an introduction to what they are. We did not once go through how to actually solve them and I've looked through the entirety of the literature and future classes. There's nothing regarding PDEs.

I tried to solve it on my own, but didn't get far. I though maybe using the chain rule and comparing the sides would lead somewhere, or perhaps choosing $a$ so the problem become simpler. But I get stuck no matter what. I found a few articles explaining how to solve homogenous cases, but they seem advanced as they contain methods I'm not at all familiar with. Finding anything regarding nonhomogeneous cases is even more difficult. In any case, I would very much like to understand how to solve such a problem and not just know the answer. So in my desperation and lack of sleep I am here to ask: how does one go about solving this?

ShootinLemons
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1 Answers1

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$\frac{\partial f}{\partial x}-3x \frac{\partial f}{\partial y}=y\tag{1}$

$(1)$ is a Quasi-linear pde.

Lagrange's auxiliary equation(conversion of a quasi-linear pde to a system of odes):

$\frac{dx}{1}=\frac{dy}{-3x}=\frac{df}{y}$

On solving :$$\frac{dx}{1}=\frac{dy}{-3x}$$

$$y+\frac{3x^2}{2}=C_1\tag{2}$$

On solving :

$\begin{align}&\frac{dx}{1}=\frac{df}{y}\\&(C_1-\frac{3x^2}{2})dx=df\\&C_1x-\frac{x^3}{2}+C_2=f\\&xy+x^3+C_2=f\tag{3}\end{align}$

Hence solution is given by

$\begin{align}f(x,y)-xy-x^3=\phi(y-\frac{3x^2}{2})\end{align}$

Putting initial condition:

$x^3+x^2=f(x, 0) =x^3+\phi(-\frac{3x^2}{2}) $

Hence $\phi(t) =-\frac{2t}{3}$

Complete solution:

$f(x, y) =xy+x^3+x^2-\frac{2y}{3}$

Sourav Ghosh
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  • From the links you sent it says "The general solution to a quasi-linear PDE can be written in the form $F(\psi_1,\psi_2)=0$. How did you get $\phi\left(y-\frac{3x^2}{2}\right)$? – ShootinLemons Apr 12 '23 at 21:32
  • Or $\psi_1=F(\psi_2) $ – Sourav Ghosh Apr 13 '23 at 02:29
  • The correct answer was $xy+x^3+x^2+\frac{2y}{3}$, but your answer was extremely helpful nonetheless. I think you accidently wrote a minus instead of a plus inside $\phi$ after $(3)$.

    I still don't quite understand how we get that $C_2$ is a function of $C_1$ i.e. $C_2 = \phi(C_1)$ or as you put it $\psi_1 = F(\psi_2)$. I didn't see that anywhere from the links you sent. Is it somehow implied from $F(\psi_1,\psi_2)=0$?

     – ShootinLemons Apr 16 '23 at 21:54
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    $\psi_1(x, y) =C_1$ and $\psi_2(x, y) =C_2$ $\$Any relation between $\psi_1, \psi_2(\text{ or} C_1, C_2) $ is a solution ( called integral surface) – Sourav Ghosh Apr 17 '23 at 02:21