Consider the initial value problem $\displaystyle x''+x+\varepsilon x=0$ with $\displaystyle x(0)=1, x'(0)=0$. I need to find the exact solution to this problem. Thanks.
I assume $\displaystyle \epsilon$ is constant? Then
$\displaystyle x'' + (1 + \epsilon)x = 0$
The characteristic equation is $\displaystyle m^2 + (1 + \epsilon) = 0$, so
$\displaystyle m = \pm i \sqrt{1 + \epsilon}$. So the general solution is:
$\displaystyle x(t) = Ae^{it\sqrt{1 + \epsilon}} + Be^{-it\sqrt{1 + \epsilon}}$
A physicist would probably prefer the equivalent expression:
$\displaystyle x(t) = Csin(t\sqrt{1 + \epsilon} ) + Dcos(t \sqrt{1 + \epsilon})$
Now use the form of your choice and apply the initial conditions.
-Dan
Given the ODE $\displaystyle a_ny^{(n)} + a_{n-1}y^{(n-1)} + ... + a_1y' + a_0 = 0$ (with constant coefficients) you may construct a general solution by solving the polynomial equation:
$\displaystyle a_nm^n + a_{n-1}m^{n-1} + ... + a_1m + a_0 = 0$
(the characteristic or auxiliary equation)
Solutions to the ODE will be:
$\displaystyle y = Ae^{m_1x} + Be^{m_2x} + ... + Ze^{m_nx}$
where the $\displaystyle \{ m_i \}$ are the solutions to the polynomial and A, B, ...Z are constants of integration.
-Dan