Solution to Differential Equation: u=x+y, y=0

[jacobrhcp]

[SOLVED] differential equation

Homework Statement

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

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?

 

[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?

 

[jacobrhcp]

what does the O in ODE mean?

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

 

[arildno]

O-Ordinary.

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

 

[Rainbow Child]

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

 

[Rainbow Child]

Ouppps! arildno was faster!

 

[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 separated variables have you?

 

[arildno]

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 separated variables have you?

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.

 

[jacobrhcp]

you're right. I'm mistaken

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

 

[HallsofIvy]

Homework Statement

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

The Attempt at a Solution

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

1.

$$u=x+y, \partial u = \partial y,$$
$$\frac{\partial u}{\partial x}=u^{2},$$

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.

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

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?

 

[jacobrhcp]

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