Here is a recoding of a solution to a system of linear, first-order differential equations with a slope field and solution curves near eigenvectors. or get it on You Tube, http://youtu.be/A-RwYm2SLiA Here is a recording of a similar system but with complex eigenvalues: http://youtu.be/q9nCdpdlqE4

Here is the third test. I have one more comprehensive part to post: coming soon. For now enjoy part 1 of the last test: mth256_exam3_summer12

\[\ddot{x}+x=0.5\cos(0.8x)\]

The homogeneous solution is\[x_h(t)=c_1\cos(t)+c_2\sin(t)\]

Using variation of parameters we get a nonhomogeneous solution of\[x_p(t)=\left(\frac{5}{4}\cos(0.2t)+\frac{5}{36}\cos(1.8t)\right)\cos(t)+\left(\frac{5}{4}\sin(0.2t)+\frac{5}{36}\sin(1.8t)\right)\sin(t)\]

Distributing and regrouping\[x_p(t)=\frac{5}{4}\cos(t)\cos(0.2t)+\frac{5}{4}\sin(t)\sin(0.2t)+\frac{5}{36}\cos(1.8t)\cos(t)+\frac{5}{36}\sin(1.8t)\sin(t)\]

Reversing the cosine angle difference formula\[x_p(t)=\frac{5}{4}\cos(0.8t)+\frac{5}{36}\cos(0.8t)=\frac{25}{18}\cos(0.8t)\]

Using the initial conditions \(x(0)=0\), \(\dot{x}(0)=0\), we get \(c_1=\frac{-25}{18}\) and \(c_2=0\) So a specific solution is\[x(t)=\frac{-25}{18}\cos(t)+\frac{25}{18}\cos(0.8t)\]

\[x(t)=\frac{25}{18}\left(\cos(0.8t)-\cos(t)\right)\]

Regrouping and applying a difference to product formula\[x(t)=\frac{25}{18}\left(2\sin\left(\frac{1-0.8}{2}\right)\sin\left(\frac{1+0.8}{2}\right)\right)\]

[…]