# Initial value problem for a linear system.

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

$$\frac{dY}{dt} = \begin{pmatrix} 1 & -1 \\ 1 & 3 \\ \end{pmatrix} Y \text{, where } Y(0) = \begin{pmatrix} 0 \\ 2 \\ \end{pmatrix}$$

• The initial values are $(x(0),y(0))^T$. Check that $t \mapsto e^{2t} (1,-1)^T$ is also a solution. Combine these two together appropriately. – copper.hat Mar 13 '15 at 20:06
• Why your solution has no integration constants? – Emilio Novati Mar 13 '15 at 20:08
• I've made an edit to the initial value, sorry for the confusion. – Chris Mar 13 '15 at 20:16
• it will make it a lot easier, now that you see $e^{2t}$ in both, to make a change of variable $u = ye^{-2t}, y = ue^{2t}$ – abel Mar 15 '15 at 22:11

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.

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}

• And B and D are? – Did Mar 15 '15 at 21:50
• @Did I forgot to come back and finish the solution presented. – Leucippus Mar 15 '15 at 22:22

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

• This is correct but it amounts to drawing the formula for $e^{At}$ from a hat, no? – Did Mar 15 '15 at 21:52
• @did: I added a link. In my opinion the exponential of a matrix should be an essential part of a course in linear differential equations. And for $2\times2$ matrices it is easy. – Emilio Novati Mar 16 '15 at 8:53
• Sure (and not only for maths students...), but would anybody mastering it post such a question? – Did Mar 16 '15 at 8:59