Solving a matrix initial value problem

Question

Let $A$ be a $2 \times 2$ matrix with $-3$ and $-1$ as eigenvalues. The eigenvectors are $v_1=[-1,1]$ and $v_2=[1,1]$. Let $x(t)$ be the position of a particle at time $t$. Solve the initial value problem $x'(t)=Ax$, $x(0)=[2,3]$.

So this should be easy, we set up the system as two ODEs:

\begin{align} x'(t) &= -e^{-3t}+e^{-t} \\ y'(t) &= e^{-3t}+e^{-t} \end{align}

Integrating each eqn.:

\begin{align} x(t) &= 3e^{-3t}-e^{-t} \\ y(t) &=-3e^{-3t}-e^{-t} \end{align}

But without the coefficients there is nothing to discover. So I insert the coefficients:

\begin{align} x(t) &= C_13e^{-3t}-C_2e^{-t} \\ y(t) &=-C_13e^{-3t}-C_2e^{-t} \end{align}

and apply the initial conditions:

\begin{align} 2 &= 3C_1-C_2 \\ 3 &=-3C_1-C_2 \end{align}

but this gives:

\begin{align} \frac{2+C_2}{3} &= C_1 \\ 3 &= -2 \end{align}

which is clearly incorrect.

Any ideas?

Thanks

Answer 1

You misunderstood something from the start. The matrix $A$ is not explicitly given. From what is known of the general solution, you get $$ \pmatrix{x\\y}=c_1v_1e^{-3t}+c_2v_2e^{-t}. $$ There is no further differentiation or integration required. Just the solution of the linear system for the initial values $$ \pmatrix{2\\3}=c_1\pmatrix{-1\\1}+c_2\pmatrix{1\\1} \\~\\ 2=-c_1+c_2\\ 3=c_1+c_2 $$ which should be easily found as $c_1=0.5$ and $c_2=2.5$

Answer 2

You made a mistake when you solved the linear system. Adding the two equations gives $5 = -2C_2$, hence $C_2 = -\frac{5}{2}$. Plugging this into one of the equations, we obtain $C_1 = -\frac{1}{6}$.

Answer 3

Where you claim you “integrated each eqn.”, you actually differenciated them instead:

\begin{align} x(t) &= \frac13 e^{-3t}-e^{-t} \\ y(t) &=-\frac13 e^{-3t}-e^{-t} \end{align}

The system for the constants after applying the initial conditions becomes:

\begin{align} 2 &= \frac13 C_1-C_2 \\ 3 &=-\frac13 C_1-C_2 \end{align}

Add both to get $5=-2C_2$, then substract the second from the first to get $-1=\frac23 C_1$. Hence, $$C_1 = -\frac32, \quad C_2 = -\frac52.$$