How Do I Solve This Differential Equation System?

[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!

 

[berkeman]

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!

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

 

[ForMyThunder]

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

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

The book is here:

http://books.google.com/books?id=NI...rontcover&dq=differential+equations#PPA259,M1

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. =\

 

[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 ?

 

[ForMyThunder]

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 ?

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.

 

[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 :(

 

[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