Can You Solve This Linear System of ODE with Initial Conditions?

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Here is this week's POTW:

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Solve the linear system of ODE

\[\begin{align}
\frac{dx}{dt} = 3x + 4y\\
\frac{dy}{dt} = 4x - 3y
\end{align}\]

with initial conditions $x(0) = 1$, $y(0) = 0$.

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This week's problem was solved correctly by Ackbach. Honorable mention goes to Cbarker1 for a partially correct solution. You can read Ackbach's solution below.

Spoiler

Let
$$A=\left[\begin{matrix}3&4\\4&-3\end{matrix}\right]\quad\text{and}\quad\mathbf{x}=\left[\begin{matrix}x\\y\end{matrix}\right]
\quad\text{and}\quad\mathbf{x}_0=\left[\begin{matrix}1\\0\end{matrix}\right].$$
Then the system is $\dot{\mathbf{x}}=A\mathbf{x}.$ The solution is generally $\mathbf{x}=e^{At}\mathbf{x}_0.$ So we must calculate $e^{At}.$ There are many ways to exponentiate a matrix, but the usual way, if possible, is to diagonalize the matrix $A$ as follows:
\begin{align*}
\det(A-\lambda I)&=0\\
\left|\begin{matrix}3-\lambda&4\\4&-3-\lambda\end{matrix}\right|&=0 \\
(3-\lambda)(-3-\lambda)-16&=0 \\
\lambda^2-25&=0 \\
(\lambda-5)(\lambda+5)&=0 \\
\lambda&=\pm 5.
\end{align*}
Now for the corresponding eigenvectors:
\begin{align*}
\left[\begin{matrix}3-5&4\\4&-3-5\end{matrix}\right]\chi_5&=0 \\
\left[\begin{matrix}-2&4\\4&-8\end{matrix}\right]\chi_5&=0 \\
\left[\begin{matrix}-1&2\\0&0\end{matrix}\right]\chi_5&=0 \\
\chi_5&=s\left[\begin{matrix}2\\1\end{matrix}\right] \\
&=\frac{1}{\sqrt{5}}\left[\begin{matrix}2\\1\end{matrix}\right] \quad\text{(normalized);}\\
\left[\begin{matrix}3+5&4\\4&-3+5\end{matrix}\right]\chi_{-5}&=0 \\
\left[\begin{matrix}8&4\\4&2\end{matrix}\right]\chi_{-5}&=0 \\
\left[\begin{matrix}2&1\\0&0\end{matrix}\right]\chi_{-5}&=0 \\
\chi_{-5}&=s\left[\begin{matrix}1\\-2\end{matrix}\right] \\
&=\frac{1}{\sqrt{5}}\left[\begin{matrix}1\\-2\end{matrix}\right].
\end{align*}
Note that $\chi_5\cdot\chi_{-5}=0,$ and $\chi_5\cdot\chi_5=1,$ and $\chi_{-5}\cdot\chi_{-5}=1.$ So this is an orthonormal set, and since we found two such vectors, the matrix is diagonalizable. We have
$$A=\frac{1}{5}\left[\begin{matrix}2&1 \\1 &-2 \end{matrix}\right]\left[\begin{matrix}5&0 \\0 &-5 \end{matrix}\right]\left[\begin{matrix}2&1 \\1 &-2 \end{matrix}\right]=\frac15\left[\begin{matrix}2&1 \\1 &-2 \end{matrix}\right]\left[\begin{matrix}10&5 \\-5 &10 \end{matrix}\right]
=\frac15\left[\begin{matrix}15&20 \\20 &-15 \end{matrix}\right]=A, $$
as required. We let
$$D=\left[\begin{matrix}5&0 \\0 &-5 \end{matrix}\right],\quad Q=\frac{1}{\sqrt{5}}\left[\begin{matrix}2&1 \\1 &-2 \end{matrix}\right],$$
so that $A=QDQ^{\dagger};$ since $Q^{\dagger}=Q,$ we can just write $A=QDQ.$ It follows that
\begin{align*}e^{At}&=\sum_{n=0}^{\infty}\frac{(At)^n}{n!}\\
&=\sum_{n=0}^{\infty}\frac{A^nt^n}{n!}\\
&=\sum_{n=0}^{\infty}\frac{(QDQ)^nt^n}{n!}\\
&=Q \left[\sum_{n=0}^{\infty}\frac{D^nt^n}{n!}\right]Q\\
&=Qe^{Dt}Q\\
&=\frac15\left[\begin{matrix}2&1 \\1 &-2 \end{matrix}\right]\left[\begin{matrix}e^{5t}&0 \\0 &e^{-5t} \end{matrix}\right]\left[\begin{matrix}2&1 \\1 &-2 \end{matrix}\right]\\
&=\frac15\left[\begin{matrix}2&1 \\1 &-2 \end{matrix}\right]\left[\begin{matrix}2e^{5t}&e^{5t} \\e^{-5t} &-2e^{-5t} \end{matrix}\right]\\
&=\frac15\left[\begin{matrix}4e^{5t}+e^{-5t}&2e^{5t}-2e^{-5t} \\2e^{5t}-2e^{-5t} &e^{5t}+4e^{-5t} \end{matrix}\right].
\end{align*}
Finally, we compute
$$e^{At}\mathbf{x}_0=\frac15\left[\begin{matrix}4e^{5t}+e^{-5t}&2e^{5t}-2e^{-5t} \\2e^{5t}-2e^{-5t} &e^{5t}+4e^{-5t} \end{matrix}\right]
\left[\begin{matrix}1\\0\end{matrix}\right]=\frac15\left[\begin{matrix}4e^{5t}+e^{-5t}\\2e^{5t}-2e^{-5t}\end{matrix}\right]=\mathbf{x}.$$
Just as a note, we can write this with hyperbolic trig functions as follows:
$$\mathbf{x}=\left[\begin{matrix}\cosh(5t)+3\sinh(5t)/5 \\ 4\sinh(5t)/5\end{matrix}\right].$$