# fundamental matrix with tricky integrals

I am trying to work out the fundamental matrix of the system:

x′1=−kx1x′2=he−ktx1−(2tsin2t−cos2t+2k)x2\begin{array}{ccc}
x_1′ &= & -kx_1 & \\
x_2′ & = & he^{-kt}x_1 &-(2t\sin2t -\cos 2t +2k) x_2 \\
\end{array}

So I choose solution x1=e−ktx_1=e^{-kt} and the equation involving x2x_2 becomes:

x′2=he−2kt−(2tsin2t−cos2t+2k)x2x_2’=he^{-2kt}-(2t\sin2t -\cos 2t +2k) x_2

To solve this we get the integrating factor : e−tcos2t+2kte^{-t\cos 2t +2kt}

Then the solution of x2x_2 involves the integral ∫e−tcos2tdt\int e^{-t\cos 2t}dt which I believe not expressible in terms of the elementary functions.

Now I have to show that the zero solution is not stable (for 1<2k< 3/2 and h ≠\neq 0) or equivalently the fundamental matrix Φ\Phi is not bounded. So I choose the other independent solution with x1=0x_1=0 and a corresponding x2x_2 these are easy integrals. Now the unbounded property for the fundamental matrix comes down to showing the integral ∫e−tcos2tdt\int e^{-t\cos 2t}dt is roughly ete^{t}. How do I show that? Is there any other way to get the fundamental matrix? Am I missing something? ================= ================= =================