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.