# Solution of inhomogenous system with fundamental matrix.

Could anyone show me how to calculate the general solution to $$x' = \begin{pmatrix} 3 & -1 \\ 4 & -1 \end{pmatrix}x+\begin{pmatrix} 1 \\ t \end{pmatrix}$$ where $$x_1(0) = 1$$ and $$x_2(0) = 0$$?

I get the fundamental matrix $$e^{tA} = e^t+(A-I_2)te^{t}$$

And then I want to use a theorem that says that the general solution will be on the form $$x(t) = F(t)c+F(t) \int_{t_0}^{t} F^{-1}(\tau)b(\tau)d\tau$$ where $$F(t)$$ is the fundamental matrix, and c is a column-matrix dependent on the initial value given.

In this case, this would be: $$x(t) = e^{tA}\begin{pmatrix} 1 \\ 0 \end{pmatrix}+e^{tA} \int_{0}^{t} e^{-\tau A}\begin{pmatrix} 1 \\ \tau \end{pmatrix} d\tau$$

But then it feels like the computations get nasty, when you substitute $$e^t+(A-I_2)te^t$$ for $$e^{tA}$$. Is there anyway to simplify this?

• What is $A$? Also, try to be self-consistent: sometimes you write $e^{tA}$, sometimes you write $F(t)$. There are also $f(A), q(A)$ which are completely useless. We can guess what is it that you mean, but that makes your text harder to read and reduces the probability that you get a good answer. Apr 26, 2023 at 15:07
• Ops, my matrix here is not diagonalizable (I have switched two of the elements). I.e. $$A = \begin{pmatrix} 3 & -1 \\ 4 & -1 \end{pmatrix}.$$ In the example on page 10, the matrix is diagonalizable, and then there is a theorem that says that a general solution will be on the form $$x(t) = c_1e^{\lambda_1 t}v_1+\ldots + c_n e^{\lambda_n t}v_n$$ I should add though, that this is in the homogenous case, i.e. we have $$x' = Ax$$ Apr 26, 2023 at 15:31
• @Moo Note that I have switched $4$ and $3$. The one I meant is not. Sorry for the inconvenience. Apr 26, 2023 at 15:34
• @Moo I have upvoted your answer. Apr 26, 2023 at 19:47

For the Fundamental Matrix, we get

$$e^{tA} = e^t + (A-I_2)te^{t} = e^t \left(I + t \begin{pmatrix} 2 & -1 \\ 4 & -2 \\ \end{pmatrix} \right) = e^t \begin{pmatrix} 2 t+1 & - t \\ 4 t & -2t+1 \\ \end{pmatrix}$$

Can you now continue?

Update

The intermediate calculations are

• $$e^{-\tau A}\begin{pmatrix} 1 \\ \tau \end{pmatrix} = \begin{pmatrix} e^{-\tau} \tau^2+e^{-\tau} (1-2 \tau) \\ e^{-\tau} \tau (2 \tau+1)-4 e^{-\tau} \tau \\ \end{pmatrix}$$

• $$\displaystyle \int_0^t e^{-\tau A}\begin{pmatrix} 1 \\ \tau \end{pmatrix}d\tau = \begin{pmatrix} -e^{-t} t^2-e^{-t}+1 \\ -2 e^{-t} t^2-e^{-t} t-e^{-t}+1 \\ \end{pmatrix}$$

• $$\displaystyle e^{tA} \int_0^t e^{-\tau A}\begin{pmatrix} 1 \\ \tau \end{pmatrix}d\tau = \begin{pmatrix} e^t t-t+e^t-1 \\ 2 e^t t-3 t+e^t-1 \\ \end{pmatrix}$$

The final solution is

$$X(t) = \begin{pmatrix} x_1(t)\\ x_2(t) \end{pmatrix} = e^{tA}\begin{pmatrix} 1 \\ 0 \\ \end{pmatrix} + e^{tA} \int_0^t e^{-\tau A}\begin{pmatrix} 1 \\ \tau \end{pmatrix}d\tau = \begin{pmatrix} e^t (3 t+2)-t-1 \\ e^t (6 t+1)-3t -1 \\ \end{pmatrix}$$

• Very good! Yes, this looks good. Thanks! Apr 26, 2023 at 16:05
• Hm, it does not look like the answer will be correct though. The answer should be $$x(t) = e^t \begin{pmatrix} 1 \\ 2 \end{pmatrix} +te^{t} \begin{pmatrix} 3 \\ 6 \end{pmatrix} - \begin{pmatrix} t+1 \\ 3t+1 \end{pmatrix}.$$ If I use the initial value $$x(0) = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$$ and what you provided, we get $$x(t) = e^t\begin{pmatrix} 2t+1 \\ 4t \end{pmatrix}+e^t\begin{pmatrix} 2t+1 & -t \\ 4t & -2t+1 \end{pmatrix} \int_{0}^{t} e^{-\tau}\begin{pmatrix} -2\tau+1 & \tau \\ -4\tau & 2\tau+1 \end{pmatrix} \begin{pmatrix} 1 \\ -\tau \end{pmatrix} d\tau$$ . Apr 26, 2023 at 16:27
• Thanks for checking @Moo Apr 26, 2023 at 18:54
• Yes, I discovered my mistake, I had mistakenly written $$b(\tau) = \begin{pmatrix} 1 \\ -\tau \end{pmatrix}.$$ Should not be a minus in front of $\tau$ there. Apr 26, 2023 at 20:51
• Yep, I get the same answer as you now, so it is quite likely that it is a typo in the suggested answer from the book. Apr 26, 2023 at 21:06