# Solving a homogeneous linear PDE [closed]

I am trying to solve the following PDE:

$$\frac{\partial F(x,y) }{\partial x} (y-y^3) +\frac{\partial F(x,y)}{\partial y} (-x-y^2) = 0$$ However I am not sure how to succeed. I have tried guessing functions for $F$ but nothing seems to be working. How can solve this PDE?

## closed as off-topic by Namaste, Claude Leibovici, user91500, B. Goddard, GlorfindelJul 20 '17 at 13:50

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Namaste, Claude Leibovici, user91500, B. Goddard, Glorfindel
If this question can be reworded to fit the rules in the help center, please edit the question.

• With the method of characteristics, it is possible to obtain the solution on the form of complicated implicit equation. If it's an home work, this should be not of so high level. Most likely Dman made a mistake in typing the question, or don't give the exact and complete wording of the problem. Why wasting time to edit a long and complicated answer which is never going to be used ? – JJacquelin Jul 14 '17 at 17:25

Using the method of characteristics you can assume $$F(s)=F(x(s),y(s)).$$ Differentiate this with respect to $s$ to get

$$\frac{dF}{ds}=\frac{\partial F}{\partial x}\frac{dx}{ds}+\frac{\partial F}{\partial y}\frac{dy}{ds}.$$

Comparision with the initial PDE will result in three equations:

$$\dfrac{dF}{ds}=0\implies F(x(s),y(s))=c_1$$ $$\dfrac{dx}{ds}=y-y^3$$ $$\dfrac{dy}{ds}=-x-y^2.$$

The last two equations can be divided by each other to get:

$$\dfrac{dy}{dx}=\frac{x+y^2}{y^3-y}.$$

And this ODE does not have a simple solution (use Wolfram alpha/Mathematica/Maple/Sympy)

$$-2/3\,\sqrt {3}{\rm arctanh} \left(1/3\,{\frac { \left( \left( y \left( x \right) \right) ^{2}-x-2 \right) \sqrt {3}}{x+1}}\right)+ \ln \left( \left( y \left( x \right) \right) ^{4}-2\,x \left( y \left( x \right) \right) ^{2}-2\,{x}^{2}-4\, \left( y \left( x \right) \right) ^{2}-2\,x+1 \right) -c_1=0 .$$ It is very unlikely that you will obtain a closed form soltuion to this equation. If this is a homework example you should check, if you wrote down the correct PDE.

• Mr.YouMath. I inadvertantly downvoted your question earlier today, When I reviewed my actions on the site today, thus far, I saw that my vote had been "down" instead of "up". Any way, I made a very minor edit, so I could undo the downvote and replace it with an upvote! cheers. – Namaste Jul 19 '17 at 18:45

$$\frac{\partial F(x,y) }{\partial x} (y-y^3) +\frac{\partial F(x,y)}{\partial y} (-x-y^2) = 0$$ Solving with the method of characteristics.

Set of ODEs for the characteristic curves : $\quad \frac{dx}{y-y^3}=\frac{dy}{-x-y^2}=\frac{dF}{0}$

A first family of characteristic curves comes from $dF=0 \quad\to\quad F=c_1$

A second family of characteristic curves comes from $\quad \frac{dx}{y-y^3}=\frac{dy}{-x-y^2}$

With $\begin{cases} X=x+1\\ Y=y^2-1 \end{cases} \quad\to\quad 2\frac{dX}{Y}=\frac{dY}{X+Y}\quad$: Homogeneous ODE.

Change of function : $Y(X)=XU(X) \quad\to\quad \frac{dX}{X}=\frac{UdU}{2-2U-U^2}$

Integration and a few calculus lead to : $\quad (U+\sqrt{3}+1)^{\frac{3+\sqrt{3}}{2}}(-U+\sqrt{3}-1)^{\frac{3-\sqrt{3}}{2}}X=c_2$

$$(\frac{y^2}{x+1} +\sqrt{3}+1)^{\frac{3+\sqrt{3}}{2}}(-\frac{y^2}{x+1}+\sqrt{3}-1)^{\frac{3-\sqrt{3}}{2}}(x+1)=c_2$$

The general solution of the PDE is : $$F(x,y)=\Phi\left((\frac{y^2}{x+1} +\sqrt{3}+1)^{\frac{3+\sqrt{3}}{2}}(-\frac{y^2}{x+1}+\sqrt{3}-1)^{\frac{3-\sqrt{3}}{2}}(x+1) \right)$$ where $\Phi$ is any differentiable function$. Some boundary conditions have to be specified if we expect to determine the function$\Phi$.$\dfrac{\partial F(x,y)}{\partial x}(y-y^3)+\dfrac{\partial F(x,y)}{\partial y}(-x-y^2)=0\dfrac{\partial F(x,y)}{\partial x}+\dfrac{x+y^2}{y^3-y}\dfrac{\partial F(x,y)}{\partial y}=0$Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:$\dfrac{dx}{dt}=1$, letting$x(0)=0$, we have$x=t\dfrac{dy}{dt}=\dfrac{x+y^2}{y^3-y}=\dfrac{t+y^2}{y^3-y}$Let$u=y^2-1$, Then$\dfrac{du}{dt}=2y\dfrac{dy}{dt}\therefore\dfrac{1}{2y}\dfrac{du}{dt}=\dfrac{t+y^2}{y^3-y}\dfrac{du}{dt}=\dfrac{2t+2y^2}{y^2-1}\dfrac{du}{dt}=\dfrac{2t+2u+2}{u}\dfrac{du}{dt}=\dfrac{2t+2}{u}+2$Let$v=\dfrac{u}{t+1}$, Then$u=(t+1)v\dfrac{du}{dt}=(t+1)\dfrac{dv}{dt}+v\therefore(t+1)\dfrac{dv}{dt}+v=\dfrac{2}{v}+2(t+1)\dfrac{dv}{dt}=\dfrac{2}{v}+2-v(t+1)\dfrac{dv}{dt}=-\dfrac{v^2-2v-2}{v}\dfrac{v}{v^2-2v-2}~dv=-\dfrac{dt}{t+1}\int\dfrac{v}{(v-1)^2-3}~dv=-\int\dfrac{dt}{t+1}\int\dfrac{v}{(v-1+\sqrt3)(v-1-\sqrt3)}~dv=-\int\dfrac{dt}{t+1}\int\left(\dfrac{3+\sqrt3}{6(v-1+\sqrt3)}+\dfrac{3-\sqrt3}{6(v-1-\sqrt3)}\right)~dv=-\int\dfrac{dt}{t+1}\dfrac{(3+\sqrt3)\ln(v-1+\sqrt3)}{6}+\dfrac{(3-\sqrt3)\ln(v-1-\sqrt3)}{6}=-\ln(t+1)+c_1(3+\sqrt3)\ln(v-1+\sqrt3)+(3-\sqrt3)\ln(v-1-\sqrt3)+6\ln(t+1)=c_2(v-1+\sqrt3)^{3+\sqrt3}(v-1-\sqrt3)^{3-\sqrt3}(t+1)^6=y_0\left(\dfrac{y^2-1}{t+1}-1+\sqrt3\right)^{3+\sqrt3}\left(\dfrac{y^2-1}{t+1}-1-\sqrt3\right)^{3-\sqrt3}(t+1)^6=y_0(y^2-1+(\sqrt3-1)(t+1))^{3+\sqrt3}(y^2-1-(\sqrt3+1)(t+1))^{3-\sqrt3}=y_0(y^2-1+(\sqrt3-1)(x+1))^{3+\sqrt3}(y^2-1-(\sqrt3+1)(x+1))^{3-\sqrt3}=y_0\dfrac{dF}{dt}=0$, letting$F(0)=f(y_0)$, we have$F(x,y)=f(y_0)=f((y^2-1+(\sqrt3-1)(x+1))^{3+\sqrt3}(y^2-1-(\sqrt3+1)(x+1))^{3-\sqrt3})\$