# Differential equation

1. Feb 16, 2008

### jacobrhcp

[SOLVED] differential equation

1. The problem statement, all variables and given/known data

$$\frac{\partial y}{\partial x}=(x+y)^{2}, y(0)=0$$

3. The attempt at a solution

I have made two attempts, both using the same substitution, where I think I made an error.

1.

$$u=x+y, \partial u = \partial y,$$
$$\frac{\partial u}{\partial x}=u^{2},$$
$$-\frac{1}{u}=x+c_{0},$$
$$y=\frac{x^{2}+c_{0}+1}{x}, y(0)=0,$$
$$y=-x$$

checking the solution gives -1=0, which is false.

2.

$$u=x+y, \partial u = \partial y,$$
$$\frac{\partial u}{\partial x}=1,$$
$$(x+y)=x+c_{0},$$
$$y=c_{0}=0$$,

which gives 0=x^{2} after putting it back in the differential equation.

I'm sure I did something fundamentally wrong, and I don't know why I would be allowed to make these substitutions. I did them because every other way of solving this led to something I didn't believe. Could anyone enlighten me on the idea of substitution?

Last edited: Feb 16, 2008
2. Feb 16, 2008

### Rainbow Child

The substitution $u(x)=y(x)+x$ gives $u'(x)=1+y'(x)$, thus your ODE reads $u'(x)=u(x)^2+1$ which is a separable one. Can you solve this one?

3. Feb 16, 2008

### jacobrhcp

what does the O in ODE mean?

and how would you seperate your variables then? because u is dependent of x, I don't know what you're allowed to do next.

4. Feb 16, 2008

### arildno

O-Ordinary.

You may write:
$$\frac{du}{u^{2}+1}=dx$$

5. Feb 16, 2008

### Rainbow Child

Ordinary Differential Equation = ODE.
Just write
$$u'=u^2+1\Rightarrow \frac{d\,u}{u^2+1}=d\,x$$
and integrate it.

6. Feb 16, 2008

### Rainbow Child

Ouppps! arildno was faster!

7. Feb 16, 2008

### jacobrhcp

okay I can do that, so thanks a lot :)

but why are you allowed to integrate, because when u is dependent of x, you haven't really made seperated variables have you?

8. Feb 16, 2008

### arildno

Okay; here you are mixing together the "separation of variables" technique in partial differential equations, and the chain rule trick in one-variable diff. eqs called "separation of variables".

In partial diff.eqs, the "separation" has nothing to do with the chain rule of differentiation, but by ASSUMING a solution u(x,y)=F(x)*G(Y), i.e, that a solution u can be found by writing it as a product of two single-variable functions.

Last edited: Feb 16, 2008
9. Feb 16, 2008

### jacobrhcp

you're right. I'm mistaken

and thanks. I solved the problem (well, you helped too)... ;)

10. Feb 16, 2008

### HallsofIvy

Staff Emeritus
Don't use "curly d"s, these are not partial derivatives. In particular, x and y are not independent variables. The whole point of the equation is that y is a function of x!
If u= x+ y, then du/dx= 1+ dy/dx.
Using that, your equation becomes du/dx- 1= u2 so solve du/dx= u2+ 1.

11. Feb 16, 2008

### jacobrhcp

wow, you're all a nice help =)... duely noted and appreciated.