# system of ode with non-constant coefficient matrix

I am sorry but I haven't learn any method to solve this kind of problem if the given matrix is non-constant.

$$\begin{pmatrix}x\\y\end{pmatrix}^\prime=\begin{pmatrix} 1&-\cos t \\ \cos t & 1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}$$

Thanks

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What you need is the matrix exponential. In a linear system of ODEs $\dot{x}=A(t)x$, if the matrix function commutes with itself over any two arguments, i.e. if $A(t)A(s)=A(s)A(t)$ for all $t,s$, then we can write $B(t)=\int_0^tA(\tau)d\tau$ and observe $B(t)A(s)=A(s)B(t)$ hence $\frac{d}{dt}B(t)^n=B'(t)B(t)^{n-1}$, which we can then use to evaluate the derivative of $X(t)=\exp B(t)$ (ignoring uniformity issues) $$\frac{d}{dt}X(t)= \frac{d}{dt}\sum_{n=0}^\infty\frac{B(t)^n}{n!}=\sum_{n=0}^\infty B'(t)\frac{B(t)^n}{n!}=B'(t)\exp B(t)=A(t)X(t).$$ Hence the solution to $\dot{x}=A(t)x$ under these condtions is $x(t)=X(t)x_0$, since $X(0)=I$ and differentiation won't be bothered by the vector multiplication (technically there are still existence/uniqueness theorems lurking in the background, keeping this rigorous).

You can check that in your case the matrix $A(t)$ does indeed commute with itself over any two arguments, hence we should compute the matrix exponential. Write $u=\sin t$, then use diagonalization for a fast route (using $\exp(PAP^{-1})=P(\exp A)P^{-1}$):

$$\exp\begin{pmatrix}t&-u\\u&t\end{pmatrix} =\exp\left[\frac{1}{\sqrt{2}}\begin{pmatrix}-i&i\\1&1\end{pmatrix}\cdot\begin{pmatrix}t-iu&0\\0&t+iu\end{pmatrix}\cdot\frac{1}{\sqrt{2}}\begin{pmatrix}i&1\\-i&1\end{pmatrix}\right]$$

$$=\frac{1}{\sqrt{2}}\begin{pmatrix}-i&i\\1&1\end{pmatrix}\begin{pmatrix}e^{t-iu}&0\\0&e^{t+iu}\end{pmatrix}\frac{1}{\sqrt{2}}\begin{pmatrix}i&1\\-i&1\end{pmatrix} =e^t\begin{pmatrix}\cos u&-\sin u\\\sin u&\cos u\end{pmatrix}.$$

(Or we could have cheated with W|A.) Note that exponentiation of diagonal matrices is straightforward; it is easy to check via power series that $$\exp\;\mathrm{diag}(\lambda_1,\dots,\lambda_n)=\mathrm{diag}(e^{\lambda_1},\dots,e^{\lambda_n}).$$ Thus our solution is $$\begin{pmatrix}x(t)\\y(t)\end{pmatrix}=e^t\begin{pmatrix}\cos(\sin t)&-\sin(\sin t)\\\sin(\sin t)&\cos(\sin t)\end{pmatrix}\begin{pmatrix}x_0\\y_0\end{pmatrix}.$$

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Yes, matrix exponential is the staff that I am learning now. However, it is hard for me to relate this knowledge to solve this question. Big thanks, anon. – Qomo Sep 29 '11 at 4:06
@Qomo: My answer here was pretty quick-paced (and the first part is just background about when the solution method is valid). If you have trouble wading through it I'll see if I can thin it out a bit. – anon Sep 29 '11 at 4:09