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 ?

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.

[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(?)

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.