allen_83

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. how ?

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'' = ex -1. The general answer to the equation will be the sum of inhom. + hom. solutions of the equation. Any help is appreciated.