# Construct a differential equation from the basis of solution

1. Dec 8, 2016

### Mr Davis 97

1. The problem statement, all variables and given/known data
Write down a 3x3 matrix A such that the equation $\vec{y}'(t) = A \vec{y}(t)$ has a basis of solutions $y_1=(e^{-t},0,0),~~y_2 = (0,e^{2t},e^{2t}),~~y_3 = (0,1,-1)$

2. Relevant equations

3. The attempt at a solution
I was thinking that, it looks like the matrix would have to have eigenvalues -1, 2, 0, with corresponding eignevectors (1,0,0), (0,1,1), and (0,1,-1). However, I am not sure how I would use this information to construct a matrix.

Last edited: Dec 8, 2016
2. Dec 8, 2016

### Ray Vickson

Are you sure that the third basis member is $y_3 = (0,1,-1)?$ I ask because $y_3\prime \equiv (0,0,0)$ and that is not a non-zero linear combination of $y_1, y_2, y_3$, which means that the problem looks impossible.

3. Dec 8, 2016

### Mr Davis 97

Yes, I am sure. One answer to the question that the author gives is $\begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 1 \end{bmatrix}$, if that helps.

4. Dec 8, 2016

### Staff: Mentor

If we look at the solutions, we'll have again two blocks here: the first coordinate with the solution $\vec{y}_1$ and the other two with $\vec{y}_2,\vec{y}_3$. Let's write $\vec{y}(t)=(x_1(t),x_2(t),x_3(t))$ to not confuse the coordinates of $\vec{y}(t)$ with the numbering of the solutions.

So we have in the first coordinate the equation $x_1'(t)=a_{11}x_1(t)$ which is solved by $x_1(t)=\exp(a_{11}t)+C$. Since $x_1(t)=(\vec{y}_1(t))_1=\exp(-t)$ is a given solution, we can chose $a_{11}=-1$ and $C=0$.

Thus we are left with $\vec{y}'(t)=\begin{bmatrix}-1&0\\0&A\end{bmatrix}\cdot \vec{y}(t)$ or $\begin{bmatrix}x_2'(t)\\x_3'(t)\end{bmatrix}=A\cdot \begin{bmatrix}x_2(t)\\x_3(t)\end{bmatrix} = \begin{bmatrix}a&b\\c&d\end{bmatrix}\cdot \begin{bmatrix}x_2(t)\\x_3(t)\end{bmatrix}$.

Now we have to solve these two equations, compare the solutions with $y_2(t)=(e^{2t},e^{2t})$ and $y_3(t)=(1,-1)$ to determine $a,b,c,d$ the way we did with $a_{11}=-1$ in the first coordinate.
(I haven't done it. It's just how I would approach the problem.)

5. Dec 9, 2016

### pasmith

The first column of the matrix is the image of (1,0,0). The second column is the image of (0,1,0). The third column is the image of (0,0,1).

Now you already know the image of (1,0,0), and from the known images of (0,1,1) and (0,1,-1) you can find the images of (0,1,0) and (0,0,1).