Recent content by Iacha

If you know the homogeneous solutions, I believe you can get the particular one you need by applying the method of variation of parameters which works even in the non constant coefficient case :) http://en.wikipedia.org/wiki/Variation_of_parameters

I did again the calculation and this time I think I got it right, Starting from the test solution: \begin{equation}u(x)=x(A\cos(2x)+B\sin(2x))\end{equation} by applying the usual rules of differentiation one obtains: \begin{equation}u(x)''= 0(A\cos(2x)+B\sin(2x)) +...

Hi, everyone! This is my first post here, I need an hand with this equation! Homework Statement Solve the initial value problem: \begin{equation} \begin{cases} u''(x)+4u(x)=\cos(2x) \\u(0)=u'(0)=1 \end{cases} \end{equation} The Attempt at a Solution I started by solving the...