# Help with differential equation

1. Aug 5, 2008

### ForMyThunder

I've been having trouble with this one differential equation for a VERY long time. This is not homework.

$$\frac{dx}{y+z}$$ = $$\frac{dy}{x+z}$$ = $$\frac{dz}{x+y}$$

Any suggestions on where to start? Any advice will be much appreciated.

Thanks!

Last edited: Aug 5, 2008
2. Aug 5, 2008

### Staff: Mentor

Is it still coursework? If so, I need to move this to the Homework Help forums.

3. Aug 5, 2008

### ForMyThunder

No, it's in a book I'm using to learn how to solve differential equations; a century-old book.

The book is here:

On page 259, problem 3. The solution is with it.

I just need to see how to arrive at that answer. It's been bugging me for a month or two. =\

Last edited: Aug 5, 2008
4. Aug 5, 2008

### Renmazuo

Hey,

Since I can't quite browse the book (assuming such function should be available), could you possibly post here the solution provided? Furthermore, and I ask as I'm not very acquainted with the notation used, are you asked to solve for z = z(x), y = y(x)? If that is the case, how about y = z = x ?

5. Aug 5, 2008

### ForMyThunder

The solution is:

$$\sqrt{x+y+z}$$ = $$\frac{a}{z-y}$$ = $$\frac{b}{x-z}$$

On account of the chapter, I do not think that x, y, and z are functions, more like the three interdependent variables.

6. Aug 6, 2008

### Marin

Hi there!

I'm not sure if this would help solving the system, but I'll post it :)

$$\frac{dx}{y+z}=\frac{dy}{x+z}=\frac{dz}{x+y}$$

from it, we obtain the system:

$$\frac{dy}{dx}=\frac{x+z}{y+z}$$
$$\frac{dz}{dx}=\frac{x+y}{z+y}$$

where $$y\rightarrow y(x)$$ and $$z\rightarrow z(x)$$

or

$$y'=\frac{x+z}{y+z}$$
$$z'=\frac{x+y}{z+y}$$

now, adding the two equations we get:

$$y'+z'=\frac{x+z}{y+z}+\frac{x+y}{z+y}$$ or
$$y'+z'=\frac{x+z+x+y}{y+z}$$
$$y'+z'=\frac{2x+y+z}{y+z}$$

now we define $$u\rightarrow u(x)$$, such that $$u(x)=y(x)+z(x)$$

this means tnat: $$u'(x)=y'(x)+z'(x)$$, or briefly $$u'=y'+z'$$

so our equation gets reduced to:

$$u'=\frac{2x+u}{u}$$, which is a first order nonlinear equation for $$u(x)$$

Now unfortuanltey, I have no idea how the last equation could be solved :(

7. Aug 6, 2008

### Marin

Here's also my desperate attempt to solve the remaining equation:

$$u'=\frac{2x+u}{u}$$

$$uu'=2x+u$$
$$uu'-u=2x$$

$$\displaystyle{\int}uu' dx-\displaystyle{\int}u dx=\displaystyle{\int}2x dx$$

$$\displaystyle{\int}uu' dx=u^2-\displaystyle{\int}u'u dx$$

$$\displaystyle{2\int}uu' dx=u^2$$

and hence

$$\displaystyle{\int}uu' dx=\frac{u^2}{2}$$

so plugging it in the above equation

$$\frac{u^2}{2}-\displaystyle{\int}u dx=x^2$$

$$U(x)=\displaystyle{\int}u dx=\frac{u^2}{2}-x^2$$

which is I think an integral equation

maybe someone can solve it (if it's solvable) and then resubstitute $$u=y+z$$ to obtain the solution to the system

best regards, Marin

Last edited: Aug 6, 2008