Picards Method for $\sin 2t$ This is a HW question.
Derive the Taylor series for $\sin 2t$ by applying the Picard method to the
first-order system corresponding to the second-order initial value problem
$x" = −4x; x(0) = 0, x'(0) = 2.$
I know how to generate terms of a the Picard iteration if I have been given conditions of the form : $x = x + 2; x(0) = 2$ but not sure how to go about solving the above equation.
 A: For your problem, we are given:
$\tag 1 \displaystyle x''= - 4x, x(0) = 0, x'(0) = 2 \rightarrow x(t) = \sin(2t).$
However, we are asked to apply the Picard Iteration method to the first order system.
So, we need to convert $(1)$ to a first order system, so let: $x_1 = x$ and $x_2 = x'$, thus:
$\tag 2 x'_1 = x' = x_2 ~~~ \text{and} ~~ x'_2 = x'' = -4x_1$.
This gives us the matrix: $A = \begin{bmatrix}
0 & 1 \\
-4 & 0 \\
\end{bmatrix}
$
Solving for the eigenvalues and eigenvectors of $A$, yields:
$\lambda_1 = 2i, v_1 = (-\frac{i}{2}, 1)$
$\lambda_2 = -2i, v_2 = (\frac{i}{2}, 1)$
Using the eigenvalues, eigenvectors and initial conditions, we arrive at:
$x_1(t) = \sin(2t)$ and $x_2(t) = 2\cos(2t)$. You should verify that this satisfies the system in $(2)$ and derive it!
Now, we are going to write out the Taylor series for these two since we need it for comparison purposes:
$\tag 3 \displaystyle x_1(t) = \sin(2t) = 2 t-\frac{4 t^3}{3} + \frac{4 t^5}{15} + O(t^7)$
$\tag 4 \displaystyle x_2(t) = 2\cos(2t) = 2 - 4 t^2 + \frac{4 t^4}{3} - \frac{8 t^6}{45} + O(t^7)$
Now, you actually have to perform Picard Iteration on $x_1(t)$ and $x_2(t)$, but you said you know how to do this.
Of course, if everything works out, you should see these terms looking like $(3)$ and $(4)$.
Regards
A: the basic setup is.
$X_{0}=  \begin{bmatrix}
0 \\
2 \\ \end{bmatrix}$and $A = \begin{bmatrix}
0 & 1 \\
-4 & 0 \\
\end{bmatrix}$
so
$\begin{bmatrix}
0 \\
2 \\ \end{bmatrix} + \int^{t}_{0} \begin{bmatrix}
0 & 1 \\
-4 & 0 \\ \end{bmatrix}\begin{bmatrix}
0 \\
2 \\ \end{bmatrix} = \begin{bmatrix}
2t \\
2 \\ \end{bmatrix}$
next term is
$\begin{bmatrix}
0 \\
2 \\ \end{bmatrix} + \int^{t}_{0} \begin{bmatrix}
0 & 1 \\
-4 & 0 \\ \end{bmatrix}\begin{bmatrix}
2t \\
2 \\ \end{bmatrix}=\begin{bmatrix}
2t \\
2-4t^{2} \\ \end{bmatrix}$
etc.. the top term in your $\sin(2t)$ the bottom is $\cos(2t)$
Hope you got the rest of the assignment done c ya tomorrow mate ^^
