Consider the linear system
$$ \frac{dY}{dt} = \begin{pmatrix} 1 & -1 \\ 1 & 3 \\ \end{pmatrix} Y $$
(a) Show that the function
$$ Y(t) = \begin{pmatrix} te^{2t} \\ -(t + 1)e^{2t}\\ \end{pmatrix} $$
is a solution to the differential equation.
I verified this without trouble. The second part I am stuck on:
(b) Solve the initial-value problem
$$ \frac{dY}{dt} = \begin{pmatrix} 1 & -1 \\ 1 & 3 \\ \end{pmatrix} Y \text{, where } Y(0) = \begin{pmatrix} 0 \\ 2 \\ \end{pmatrix} $$
On
Consider the differential system \begin{align} \frac{dY}{dt} = \begin{pmatrix} 1 & -1 \\ 1 & 3 \\ \end{pmatrix} Y . \end{align} Let \begin{align} Y(t) = \begin{pmatrix} a(t) \\ b(t) \end{pmatrix} \end{align} such that \begin{align} \begin{pmatrix} a^{'} \\ b^{'} \end{pmatrix} = \begin{pmatrix} 1 & -1 \\ 1 & 3 \\ \end{pmatrix} \cdot \begin{pmatrix} a \\ b \end{pmatrix}. \end{align} The set of differential equations is then given by \begin{align} a' &= a - b \\ b' &= a + 3b \end{align} of which $a$ and $b$ satisfy the second order equation $w'' - 4 w' +4 w = 0$. The solution of this second order equation is $w(t) = (A + B t ) e^{2t}$ and lead to the form of $Y(t)$ being \begin{align} Y(t) = \begin{pmatrix} A + B t \\ C + D t \end{pmatrix} \, e^{2t}. \end{align} Now the initial condition is \begin{align} Y(0) = \begin{pmatrix} 0 \\ 2 \end{pmatrix} \end{align} for which \begin{align} \begin{pmatrix} 0 \\ 2 \end{pmatrix} = \begin{pmatrix} A \\ C \end{pmatrix} \end{align} which yields the solution being \begin{align} Y(t) = \begin{pmatrix} B t \\ 2 + D t \end{pmatrix} \, e^{2t}. \end{align} Using this result to take its derivative and apply it to the differential equation it is seen that the result equations amount to $(B+D)t = -(B+2)$ and $(B+D)t = D-2$. In order for there to be solutions it is required that $B = -2$ and $D = 2$. The final solution becomes \begin{align} Y(t) = 2 \, e^{2t} \, \begin{pmatrix} - t \\ 1 + t \end{pmatrix} \end{align}
On
As @ copper.hat mentioned, you can solve this using his approach. If you want to do it from scratch, we can proceed as follows (note that there are many approaches to solving these).
The characteristic polynomial is given by $|A - \lambda I| = 0$, yielding eigenvalues:
$$\lambda^2 - 4 \lambda + 4 = 0 \implies \lambda_{1,2} = 2$$
We would then solve for eigenvectors, by solving $[A -\lambda_i I]v_i = 0$ using row-reduced-echelon-form (RREF) or whatever you prefer. The RREF we get is:
$$\begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix}v_1 = 0$$
This yields a single linearly independent eigenvector of:
$$v_1 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}$$
We need a second linearly independent eigenvector and will find a generalized one using $[A- 2I]v_2 = v_1$. The RREF for this yields:
$$ \left[\begin{array}{rrr|r} 1 & 1 & -1 \\ 0 & 0 & 0 \end{array}\right] $$
We can choose $b = 0 \implies a = - 1$, for a second linearly independent and generalized eigenvector:
$$v_2 = \begin{bmatrix} -1 \\ 0 \end{bmatrix}$$
We can now write our solution as:
$$Y(t) = c_1 e^{\lambda_1 t} v_1 + c_2 e^{\lambda_2 t}\left (t v_1 + v_2\right)$$
So, we have:
$$Y(t) = c_1 e^{2t}\begin{bmatrix} 1 \\ - 1 \end{bmatrix} + c_2 e^{2t} \left(t \begin{bmatrix} 1 \\ -1 \end{bmatrix} + \begin{bmatrix}- 1 \\ 0 \end{bmatrix} \right)$$
We have the IC $Y(0) = \begin{bmatrix} 0 \\ 2 \end{bmatrix}$, so we get:
$$c_1 = c_2 = -2$$
Our final solution is:
$$Y(t) = -2 e^{2t}\begin{bmatrix} t \\ -(t+1) \end{bmatrix} $$
Compare this to the item provided in part $a.$, but not including the IC, they are the same.
The solution of a system $$ \dfrac{d}{dt}y=Ay $$ with initial conditions $y(0)=y_0$ (where $y$ is a vector function and $A$ is a matrix) is:
$$ y= e^{At}y_0 $$ in your case we have:
$$ e^{At}= e^{2t} \begin{bmatrix} 1-t&-t\\ t&1+t \end{bmatrix} $$ so the solution is: $$ e^{At}y(0)= e^{2t} \begin{bmatrix} 1-t&-t\\ t&1+t \end{bmatrix} \begin{bmatrix} 0\\ 2 \end{bmatrix} = \begin{bmatrix} -2te^{2t}\\ 2(1+t)e^{2t} \end{bmatrix} $$
To calculate the exponetial of a matrix see the answers in: Exponential of matrix