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

In summary: 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. So, in this case, you're integrating because
  • #1
jacobrhcp
169
0
[SOLVED] differential equation

Homework Statement



[tex]\frac{\partial y}{\partial x}=(x+y)^{2}, y(0)=0 [/tex]

The Attempt at a Solution



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

1.

[tex] u=x+y, \partial u = \partial y, [/tex]
[tex] \frac{\partial u}{\partial x}=u^{2}, [/tex]
[tex] -\frac{1}{u}=x+c_{0}, [/tex]
[tex] y=\frac{x^{2}+c_{0}+1}{x}, y(0)=0, [/tex]
[tex] y=-x [/tex]

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

2.

[tex] u=x+y, \partial u = \partial y, [/tex]
[tex] \frac{\partial u}{\partial x}=1, [/tex]
[tex] (x+y)=x+c_{0}, [/tex]
[tex] y=c_{0}=0 [/tex],

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:
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  • #2
The substitution [itex]u(x)=y(x)+x[/itex] gives [itex]u'(x)=1+y'(x)[/itex], thus your ODE reads [itex]u'(x)=u(x)^2+1[/itex] which is a separable one. Can you solve this one?
 
  • #3
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.
 
  • #4
O-Ordinary.

You may write:
[tex]\frac{du}{u^{2}+1}=dx[/tex]
 
  • #5
Ordinary Differential Equation = ODE. :smile:
Just write
[tex]u'=u^2+1\Rightarrow \frac{d\,u}{u^2+1}=d\,x[/tex]
and integrate it.
 
  • #6
Ouppps! arildno was faster! :smile:
 
  • #7
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?
 
  • #8
jacobrhcp said:
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.
 
Last edited:
  • #9
you're right. I'm mistaken

and thanks. I solved the problem (well, you helped too)... ;)
 
  • #10
jacobrhcp said:

Homework Statement



[tex]\frac{\partial y}{\partial x}=(x+y)^{2}, y(0)=0 [/tex]

The Attempt at a Solution



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

1.

[tex] u=x+y, \partial u = \partial y, [/tex]
[tex] \frac{\partial u}{\partial x}=u^{2}, [/tex]
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.

[tex] -\frac{1}{u}=x+c_{0}, [/tex]
[tex] y=\frac{x^{2}+c_{0}+1}{x}, y(0)=0, [/tex]
[tex] y=-x [/tex]

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

2.

[tex] u=x+y, \partial u = \partial y, [/tex]
[tex] \frac{\partial u}{\partial x}=1, [/tex]
[tex] (x+y)=x+c_{0}, [/tex]
[tex] y=c_{0}=0 [/tex],

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?
 
  • #11
wow, you're all a nice help =)... duely noted and appreciated.
 

1. What is a differential equation?

A differential equation is a mathematical equation that relates a function to its derivatives. It shows how the rate of change of a variable is related to the value of the variable itself.

2. What are some real-world applications of differential equations?

Differential equations are used to model various natural phenomena, such as population growth, fluid dynamics, heat transfer, and electrical circuits. They are also used in engineering, economics, physics, and many other fields.

3. How do you solve a differential equation?

The methods for solving a differential equation depend on its type and complexity. Some common techniques include separation of variables, substitution, and using integral or power series solutions. Advanced methods such as Laplace transforms and numerical methods may also be used.

4. What is the difference between ordinary and partial differential equations?

An ordinary differential equation (ODE) involves only one independent variable, while a partial differential equation (PDE) involves multiple independent variables. ODEs can be solved explicitly for a single function, while PDEs involve solving for a function of several variables.

5. What are the applications of solving differential equations numerically?

Numerical methods are used to approximate solutions to differential equations when an exact solution cannot be found. This is useful in many real-world applications, such as weather forecasting, simulating physical systems, and predicting the behavior of complex systems. It also allows for the solution of differential equations that are too complex to be solved analytically.

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