We are given a system of linear differential equations: $x' = x + 2y$ $y' = x$ with initial conditions $x(0) = 0$ and $y(0) = 3$. We need to find the value of $y(2)$.

1. Problem Description

We are given a system of linear differential equations:

$x' = x + 2y$

$y' = x$

with initial conditions $x(0) = 0$ and $y(0) = 3$.

We need to find the value of $y(2)$.

2. Solution Steps

First, we can differentiate the first equation with respect to t:

$x'' = x' + 2y'$

Substitute $y' = x$ into the above equation:

$x'' = x' + 2x$

Rearrange the equation:

$x'' - x' - 2x = 0$

This is a second-order linear homogeneous differential equation with constant coefficients.

We assume a solution of the form $x(t) = e^{rt}$. Substituting this into the equation, we get the characteristic equation:

$r^2 - r - 2 = 0$

$(r-2)(r+1) = 0$

So, $r = 2$ or $r = -1$.

The general solution for $x(t)$ is:

$x(t) = c_1 e^{2t} + c_2 e^{-t}$

Now we use the initial condition $x(0) = 0$:

$x(0) = c_1 e^{0} + c_2 e^{0} = c_1 + c_2 = 0$

Therefore, $c_2 = -c_1$.

$x(t) = c_1 e^{2t} - c_1 e^{-t} = c_1(e^{2t} - e^{-t})$

Since $y' = x$, we have $y' = c_1 (e^{2t} - e^{-t})$. Integrating this with respect to t, we obtain:

$y(t) = \int c_1 (e^{2t} - e^{-t}) dt = c_1 (\frac{1}{2}e^{2t} + e^{-t}) + c_3$

Now we use the initial condition $y(0) = 3$:

$y(0) = c_1 (\frac{1}{2}e^{0} + e^{0}) + c_3 = c_1 (\frac{1}{2} + 1) + c_3 = \frac{3}{2}c_1 + c_3 = 3$

Also, from the original equation $x' = x + 2y$, when $t = 0$,

$x'(0) = x(0) + 2y(0) = 0 + 2(3) = 6$

Since $x(t) = c_1(e^{2t} - e^{-t})$, we have $x'(t) = c_1(2e^{2t} + e^{-t})$

$x'(0) = c_1(2e^0 + e^0) = 3c_1 = 6$, so $c_1 = 2$.

Substitute $c_1 = 2$ into $\frac{3}{2}c_1 + c_3 = 3$:

$\frac{3}{2}(2) + c_3 = 3$

$3 + c_3 = 3$, so $c_3 = 0$.

Therefore, $x(t) = 2(e^{2t} - e^{-t})$

$y(t) = 2(\frac{1}{2}e^{2t} + e^{-t}) = e^{2t} + 2e^{-t}$

We want to find $y(2)$:

$y(2) = e^{2(2)} + 2e^{-2} = e^4 + 2e^{-2}$

3. Final Answer

$y(2) = e^4 + 2e^{-2}$