Different methods of solving a linear system of first order DE?

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

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?

Thanks.

• 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. – 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. $$x(t)=e^{tA}x_0 = \dots = x_1\begin{pmatrix}2e^{2t}+\frac13e^{3t} \\-3e^{2t}-e^{3t} \end{pmatrix}+x_2\begin{pmatrix} -2e^{2t}-\frac23e^{3t} \\3e^{2t}+2e^{3t}\end{pmatrix}$$