$$ x'(t)=\begin{pmatrix} 1&\frac{-2}3\\3&4\end{pmatrix}x(t).$$

This is the method described in my book:

  • finding the eigenvalues of matrix $A = \begin{pmatrix} 1&\frac{-2}3\\3&4\end{pmatrix}$: $$ \begin{vmatrix} 1-\lambda&\frac{-2}3 \\3&4-\lambda\end{vmatrix}=0\iff \lambda=3, \lambda=2.$$

  • eigenvector corresponding to $\lambda=3$: $\begin{pmatrix} -1/3 \\ 1\end{pmatrix}$, eigenvector corresponding to $\lambda=2$: $\begin{pmatrix} -2/3 \\ 1\end{pmatrix}$.

  • define $C = \begin{pmatrix} -2/3 & -1/3 \\ 1&1\end{pmatrix}$, then $C^{-1} = \begin{pmatrix} -3 & 3 \\ -1&2\end{pmatrix}$, and $$ A = C\operatorname{diag}(\lambda_1,\lambda_2)C^{-1} = \begin{pmatrix} -2/3 & -1/3 \\ 1&1\end{pmatrix} \begin{pmatrix} 2&0\\0&3\end{pmatrix}\begin{pmatrix} -3 & 3 \\ -1&2\end{pmatrix}.$$

  • calculate $e^{tA} $. Using previous expression for $A$ we get $$ e^{tA}=\begin{pmatrix} -2/3 & -1/3 \\ 1&1\end{pmatrix} \begin{pmatrix} e^{2t}&0\\0&e^{3t}\end{pmatrix}\begin{pmatrix} -3 & 3 \\ -1&2\end{pmatrix} = \begin{pmatrix} 2e^{2t}+\frac13e^{3t} & -2e^{2t}-\frac23e^{3t} \\ -3e^{2t}-e^{3t} & 3e^{2t}+2e^{3t}\end{pmatrix}.$$

  • the solution of the given system, considering the initial condition $x_0=(x_1,x_2)^t$, equals to $e^{tA}x_0$: $$ e^{tA}x_0 = \dots = x_1\begin{pmatrix}2e^{2t}+\frac13e^{3t} \\-3e^{2t}-e^{3t} \end{pmatrix}+x_2\begin{pmatrix} -2e^{2t}-\frac23e^{3t} \\3e^{2t}+2e^{3t}\end{pmatrix}$$

Now, I have looked up some extra exercises online, but these seem to solve such systems in a shorter way: the solution of the system above with be given by $c_1e^{2t}\begin{pmatrix} -2/3 \\ 1 \end{pmatrix} + c_2e^{3t}\begin{pmatrix} -1/3 \\ 1\end{pmatrix}$.

What is the difference between both approaches? The method used by my book seems to have more coefficients (and is therefore maybe a little more detailed/exact?). Are these solution methods equivalent? Which one would you use?


  • 2
    $\begingroup$ Funny question. You see different types of resolutions in books, you don't tell us which, you can't tell the differences and we are deemed to guess for you. $\endgroup$ – Yves Daoust May 9 at 16:02

The advantage of the exponential matrix method is that you do not have to solve for $c_1$ and $c_2$ to satisfy the initial conditions.

You just multiply your exponential matrix by the vector of initial condition and you get your solution. $$ x(t)=e^{tA}x_0 = \dots = x_1\begin{pmatrix}2e^{2t}+\frac13e^{3t} \\-3e^{2t}-e^{3t} \end{pmatrix}+x_2\begin{pmatrix} -2e^{2t}-\frac23e^{3t} \\3e^{2t}+2e^{3t}\end{pmatrix}$$

There are other simpler ways of finding the exponential matrix using Cayley-Hamilton theorem which only uses the eigenvalues of the original matrix.

The other advantage of exponential matrix is that finding its inverse is almost trivial, so in case of solving an inhomogeneous system,by the method of variation of constants, it is very convenience.


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