# Solve first order matrix differential equation

If I have a differential equation $y'(t)=A y(t)$ where A is a constant square matrix that is not diagonalizable(although it is surely possible to calculate the eigenvalues) and no initial condition is given. And now I am interested in the fundamental matrix. Is there a general method to determine this matrix? I do not want to use the exponential function and the Jordan normal form, as this is quite exhausting. Maybe there is also an ansatz possible as it is for the special case, where this differential equation is equivalent to an n-th order ode.
I saw a method where they calculated the eigenvalues of the matrix and depending on the multiplicity n of this eigenvalue they used an exponential term(with the eigenvalue) and in each component an n-th order polynomial as a possible ansatz. Though they only did this, when they were interested in a initial value problem, so with an initial condition and not for a general solution.

I was asked to deliver an example: so $y'(t)=\begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix} y(t)$ If somebody can construct a fundamental matrix for this system, than this should be sufficient

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We have many ways to proceed and this is only a $2×2$. We can choose from:

• Nineteen Dubious Ways to Compute the Exponential of a
Matrix, Twenty-Five Years Later
• Putzer’s Method 1 and Method 2
• For non-repeated eigenvalues, we can simply write:
$$x(t) = e^{At}x_0 = Pe^{Jt}P^{-1}x_0 = c_1v_1e^{\lambda_1 t} + \ldots + c_nv_ne^{\lambda_n t}$$
• The Direct Method for repeated eigenvalues

$$\tag 1 e^{At} = \left[I+ \sum_{k=1}^\infty \dfrac{(A-\lambda I)^k}{k!}t^k\right]e^{\lambda t}$$

For the matrix $A = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}$, we have:

$\det [A-\lambda I] = \det \begin{bmatrix} 3-\lambda & -4 \\ 1 & -1-\lambda \end{bmatrix} = 0 \rightarrow \lambda^2-2 \lambda+1 = 0 \rightarrow \lambda_{1,2} = 1,1$ (a double eigenvalue). From the eigenvalues, we derive the eigenvalue/eigenvector pairs:

• $\lambda_1 = 1, v_1 = (2, 1)$
• $\lambda_2 = 1, v_2 = (1, 0)$ (the second eigenvector is a generalized one)

Lets find the matrix exponential using two different methods.

Method 1

From $(1)$, we have:

$$e^{At} = \left[I + \dfrac{(A-\lambda I)^0}{1!}t^1 \right]e^{\lambda t} = \left[\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} 2 & -4 \\ 1 & -2 \end{bmatrix}t\right] = e^{t}\begin{bmatrix} 1+2t & -4t \\ t & 1-2t \end{bmatrix}$$

Method 2

Use the Laplace Transform.

$$e^{At}=\mathcal{L}^{-1}\left((sI-A)^{-1}\right)= \mathcal{L}^{-1}\left(\begin{bmatrix}s-3 & 4 \\ -1 & s+1\end{bmatrix}^{-1}\right) = e^{t}\begin{bmatrix}1+2 t & -4 t \\ t & 1-2 t\end{bmatrix}$$

$\vdots$

Method n

Try other approaches discussed above!

Update: Method n+1

If you wanted to write $A$ using Jordan Normal Form, we would have:

$$A = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} = PJP^{-1} = \begin{bmatrix} 2 & 1 \\ 1 & 0 \end{bmatrix} \cdot \begin{bmatrix} 1 & 1 \\ 0 & 1\end{bmatrix} \cdot \begin{bmatrix} 0 & 1 \\1 & -2 \end{bmatrix}$$

To write the matrix exponential for this, we take advantage of the Jordan Block and have:

$$e^{At} = e^{PJP^{-1}t} = Pe^{Jt}P^{-1} = \begin{bmatrix} 2 & 1 \\ 1 & 0 \end{bmatrix} \cdot e^{\begin{bmatrix} 1 & 1 \\ 0 & 1\end{bmatrix}t} \cdot \begin{bmatrix} 0 & 1 \\1 & -2 \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 1 & 0 \end{bmatrix} \cdot \begin{bmatrix} e^t & te^t \\ 0 & e^t\end{bmatrix} \cdot \begin{bmatrix} 0 & 1 \\1 & -2 \end{bmatrix} = e^{t}\begin{bmatrix}1+2 t & -4 t \\ t & 1-2 t\end{bmatrix}$$

Lastly, it is worth noting that sometimes the Fundamental Matrix is given as:

$$\phi(t, t_0) = \phi(t) \cdot \phi^{-1}(t_0)$$

There are many methods for determining the matrix exponential, even for a non-diagonalizable matrix. One of the easiest is via the Laplace transform. You can check that

$$\mathcal{L}(e^{tA})(s) = (sI-A)^{-1}.$$

$$\mathcal{L}(e^{tA})(s) = \begin{bmatrix}s-3 & 4 \\ -1 & s+1\end{bmatrix}^{-1} = \frac{1}{(s-1)^2}\begin{bmatrix}s+1 & -4 \\ 1 & s-3\end{bmatrix},$$
$$e^{tA} = \begin{bmatrix}(2t+1)e^t & -4te^t \\ te^t & -(2t-1)e^t\end{bmatrix}.$$
Formally, this is only valid for $t>0$, but since the elements in $e^{tA}$ are holomorphic, the identity theorem for holomorphic functions shows that the equality is valid for all $t$.