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How to solve a question like:

Initial value problem where:

$$\frac{dY}{dt} = \left(\begin{matrix} 1 & -1 \\ 1 & 3 \end{matrix}\right) Y(t),$$ where $Y(0) = \left(\begin{matrix} 0 \\ 2 \end{matrix}\right).$

The matrix just to the left of $Y$ is a $2\times 2$ matrix. This is for Differential Equations. I don't understand how to do this besides saying $ dx/dt=x -y$, and $dy/dt = x + 3y$.

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up vote 2 down vote accepted

The method of substitution works for these. You have two equations ${dx \over dt} = x - y$ and ${dy \over dt} = x + 3y$. The first can be solved for $y$ and you get that $y = -{dx \over dt} + x$. Plugging this back into the second equation you get $$-{d^2x \over dt^2} + {dx \over dt} = x + 3(-{dx \over dt} + x)$$ This simplifies to $${d^2x \over dt^2} - 4{dx \over dt} + 4x = 0$$ You have your standard solutions $x(t) = (C_1 t + C_2)e^{2t}$ to this equation. Then $y(t) = -{dx \over dt} + x = (-C_1t + 3C_2 - C_1)e^{2t}$. Here $C_1$ and $C_2$ are arbitrary constants. In matrix form this can be written as $$\pmatrix{x(t)\cr y(t)} = C_1e^{2t}\pmatrix{t\cr -t - 1} + C_2e^{2t} \pmatrix {1 \cr 3}$$ You can get $C_1$ and $C_2$ from the initial conditions. $x(0) = 0$ leads to $C_2 = 0$, and then $y(0) = 2$ leads to $-C_1 = 2$ or $C_1 = -2$. So your solution will be $$\pmatrix{x(t)\cr y(t)} = 2e^{2t}\pmatrix{-t\cr t + 1}$$

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Find the eigenvalues of $A$ (the matrix): $(1-\lambda)(3-\lambda) +1=\lambda^2-4\lambda + 4=(\lambda -2)^2=0$. Unfortunately, if you haven't seen how to solve systems this way the repeated eigenvalue case is not a good one to start from.

An eigenvector associated to this eigenvalue is $\left(\begin{matrix} 1 \\ -1 \end{matrix}\right)$. This tells us one solution is $Y_1=\left(\begin{matrix} 1 \\ -1 \end{matrix}\right)e^{2t}$.

We have to get another linearly independent solution, and to do this we need to find a generalized eigenvector, or an $\eta$ such that $(A-2I)\eta = \left(\begin{matrix} 1 \\ -1 \end{matrix}\right)$. Such a vector is $\eta=\left(\begin{matrix} 1 \\ -2 \end{matrix}\right)$.

This gives us a second solution which is linearly independent of the form $Y_2= \left(\begin{matrix} 1 \\ -1 \end{matrix}\right)te^{2t}+\left(\begin{matrix} 1 \\ -2 \end{matrix}\right)e^{2t}$

The general solution is of the form $Y(t)=C_1 Y_1 + C_2 Y_2$. If you plug in $Y(0)=\left(\begin{matrix} 0 \\ 2 \end{matrix}\right)$ you can solve for $C_1$ and $C_2$ using standard linear algebra.

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