solve a non homogeneous differential equation

[allen_83]

I need a hint on how to solve the following DE.

x'' = e^x -1.

The general answer to the equation will be the sum of inhom. + hom. solutions of the equation.

Any help is appreciated.

[the tree]

• Actually the general sollution is the sum of the homogeneous and implicit solutions.
  
• Rearranging it like so will make it a little easier: [imath]x'' +1 = e^x[/imath]
  
• You should be able to solve the homogeneous equation [imath]x'' + 1=0[/imath] without much difficulty.
  
• Finding an implicit solution is essentially just well informed guesswork - what do you think the solution will look like?
  

[allen_83]

[math]A\cdot e^x - \dfrac {1}{2} x^2[/math] (right?)

however, considering the general form :

[math]x'' + ax' + bx = g(y)[/math]

there is no place for a constant c. So I wonder if I'm actually allowed to move -1 to the left hand side of the equation(?)

Edited March 21, 2011 by allen_83

[DrRocket]

I need a hint on how to solve the following DE.

x'' = e^x -1.

The general answer to the equation will be the sum of inhom. + hom. solutions of the equation.

Any help is appreciated.

 

Integrate twice.

[the tree]

Integrate twice.

I'm assuming you meant separation of variables, and then integrating twice, right?

[DrRocket]

I'm assuming you meant separation of variables, and then integrating twice, right?

 

Sorry, brain fart.

 

None of the techniques discussed are going to work, because this ODE is non-linear.

 

[math] \dfrac {d^2 \ x}{dt^2} - e^x = -1[/math]

 

Given the OP's statement that a solution in terms of a general solution to the homogeneous equation plus a particular solution to the inhomogeneous equation is expected (which applies to a linear ODE), I suspect that the equation written is not the equation of the real problem.

 

Should it be [math] \dfrac {d^2 \ x}{dt^2} = e^t -1[/math] ?

 

If so, then integrate twice.

[allen_83]

Sorry, brain fart.

 

None of the techniques discussed are going to work, because this ODE is non-linear.

 

[math] \dfrac {d^2 \ x}{dt^2} - e^x = -1[/math]

 

Given the OP's statement that a solution in terms of a general solution to the homogeneous equation plus a particular solution to the inhomogeneous equation is expected (which applies to a linear ODE), I suspect that the equation written is not the equation of the real problem.

 

Should it be [math] \dfrac {d^2 \ x}{dt^2} = e^t -1[/math] ?

 

If so, then integrate twice.

That's right. This ODE is non-linear.

The problem is as it is.

[math] x''(t) = e^{ x(t)} - 1 for x'(0)= 0 and x(0) = 1 [/math]

Attempts:

[math]x''(t) + 1 = e^{x(t)}; \lambda^2+1 = 0 \Rightarrow \lambda_{1} = i, \lambda_{2} = -i ; x(t) = D_{1}cos(t) + D_{2}sin(t) \because x(0) = 1 , x'(0) = 0 [/math]

solvin for D1 and D2 :

D1 = 1, D2 = 0

and pluging them back in :

[math] x(t) = cos(t)[/math]

 

[math]x_{nonhom}(t) = A\cdot e^x - \dfrac{x^2}{2}[/math]

 

[math] x'_{nonhom}(t) = A\cdot e^x - x[/math]

 

[math]x''_{nonhom}(t) = A\cdot e^x - 1 [/math]

 

[math]A\cdot e^x - 1 + 1 = e^x \Rightarrow A = 1[/math]

 

[math]x_{nonhom}(t) = e^x - \dfrac{x^2}{2}[/math]

[math] x = x_{nonhom}(t) + x_{hom}(t)[/math]

[math] x = e^{x(t)} - \dfrac{x(t)^2}{2} + cos(t) [/math]

 

correct me if I'm wrong.

Edited March 22, 2011 by allen_83

[DrRocket]

That's right. This ODE is non-linear.

The problem is as it is.

[math] x''(t) = e^{ x(t)} - 1 for x'(0)= 0 and x(0) = 1 [/math]

Attempts:

[math]x''(t) + 1 = e^{x(t)}; \lambda^2+1 = 0 \Rightarrow \lambda_{1} = i, \lambda_{2} = -i ; x(t) = D_{1}cos(t) + D_{2}sin(t) \because x(0) = 1 , x'(0) = 0 [/math]

solvin for D1 and D2 :

D1 = 1, D2 = 0

and pluging them back in :

[math] x(t) = cos(t)[/math]

 

[math]x_{Inhom}(t) = A\cdot e^x - \dfrac{x^2}{2}[/math]

[math] x'_{Inhom}(t) = A\cdot e^x - x[/math]

[math]x''_{Inhom}(t) = A\cdot e^x - 1 [/math]

[math]A\cdot e^x - 1 + 1 = e^x \Rightarrow A = 1[/math]

[math]x_{Inhom}(t) = e^x - \dfrac{x^2}{2}[/math]

[math] x(t) = x_{inhom}(t) + x_{hom}(t)[/math]

[math] x(t) = e^x - \dfrac{x^2}{2} + cos(t) [/math]

 

correct me if I'm wrong.

 

1. Since the equation is non-linear, what makes you think that the solution is expressible as the sum of "the general homogeneous solution" plus an "inhomogeneous solution" ? When the differential operator is linear, any two solutions of the inhomogeneous equation differ by a solution of the homogeneous equation. Not so for non-linear equations.

 

2. [math] x(t) = e^x - \dfrac{x^2}{2} + cos(t) [/math] fails to express x explicitly as a function of t.

 

3. [math] \dfrac {d}{dt}( e^x - \dfrac{x^2}{2} + cos(t)) = e^{x(t)} \frac {dx}{dt} -x \frac {dx}{dt} - sin(t) [/math]

 

[math] \dfrac {d^2}{dt^2}( e^x - \dfrac{x^2}{2} + cos(t))[/math][math] = e^{x(t)} (\frac {dx}{dt})^2 + e^{x(t)} \frac {d^2x}{dt^2} - (\frac {dx}{dt})^2 -x \frac {d^2x}{dt^2} - cos(t) [/math] This does not appear to help very much.

 

4. The technique you are applying in your

 

[math]x''(t) + 1 = e^{x(t)}; \lambda^2+1 = 0 \Rightarrow \lambda_{1} = i, \lambda_{2} = -i ; x(t) = D_{1}cos(t) + D_{2}sin(t) \because x(0) = 1 , x'(0) = 0 [/math]

 

is intended for homogeneous linear equations. But this is all screwed up. First because [math]x''+1=0[/math] is not homogeneous, and second because it does not help with [math]x''(t) + 1 = e^{x(t)}[/math]

 

The solutions to [math]x"=-1[/math] are [math] x(t) = - \frac {t^2}{2} - at + c [/math], so you can see that you have misapplied the characteristic equation.

[allen_83]

1. Since the equation is non-linear, what makes you think that the solution is expressible as the sum of "the general homogeneous solution" plus an "inhomogeneous solution" ? When the differential operator is linear, any two solutions of the inhomogeneous equation differ by a solution of the homogeneous equation. Not so for non-linear equations.

alright, wanted to give "the tree"'s idea a try.

got no clue. am stuck.

this is a type : [math] x'' = f(t,x,x') [/math] with both x and t missing if I were to think of it as a non-linear DE.

[math] x(t) = - \frac {t^2}{2} - at + c [/math]

how ?

Edited March 23, 2011 by allen_83

[DrRocket]

 

how ?

 

integrate twice