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The problem is to solve a system of linear differential equations using the Laplace transform:

\[
\begin{cases}
y_1' = -y_1 + 4y_2, \quad y_1(0) = 3 \\
y_2' = 3y_1 - 2y_2, \quad y_2(0) = 4
\end{cases}
\]

### Step 1: Apply the Laplace Transform
Let's denote \( \mathcal{L}\{y_1(t)\} = Y_1(s) \) and \( \mathcal{L}\{y_2(t)\} = Y_2(s) \). Recall that for any function \( y(t) \):

\[
\mathcal{L}\{y'(t)\} = sY(s) - y(0).
\]

Applying the Laplace transform to both differential equations:

1. For \( y_1' = -y_1 + 4y_2 \):
\[
sY_1(s) - y_1(0) = -Y_1(s) + 4Y_2(s).
\]
Substitute \( y_1(0) = 3 \):
\[
sY_1(s) - 3 = -Y_1(s) + 4Y_2(s).
\]
Simplifying:
\[
(s + 1)Y_1(s) - 4Y_2(s) = 3. \quad \text{(Equation 1)}
\]

2. For \( y_2' = 3y_1 - 2y_2 \):
\[
sY_2(s) - y_2(0) = 3Y_1(s) - 2Y_2(s).
\]
Substitute \( y_2(0) = 4 \):
\[
sY_2(s) - 4 = 3Y_1(s) - 2Y_2(s).
\]
Simplifying:
\[
-3Y_1(s) + (s + 2)Y_2(s) = 4. \quad \text{(Equation 2)}
\]

### Step 2: Solve the System of Equations
Let's solve these simultaneous equations to find \( Y_1(s) \) and \( Y_2(s) \).

1. Multiply Equation 1 by \( s + 2 \):
\[
(s + 2)(s + 1)Y_1(s) - 4(s + 2)Y_2(s) = 3(s + 2).
\]

2. Multiply Equation 2 by 4:
\[
-12Y_1(s) + 4(s + 2)Y_2(s) = 16.
\]

3. Adding the two equations:
\[
(s + 2)(s + 1)Y_1(s) - 12Y_1(s) = 3(s + 2) + 16.
\]
Simplify:
\[
(s^2 + 3s + 2 - 12)Y_1(s) = 3s + 6 + 16.
\]
\[
(s^2 + 3s - 10)Y_1(s) = 3s + 22.
\]
\[
Y_1(s) = \frac{3s + 22}{s^2 + 3s - 10}.
\]

### Step 3: Finding \( Y_2(s) \)
Now substitute \( Y_1(s) = \frac{3s + 22}{s^2 + 3s - 10} \) into Equation 1:
\[
(s + 1)Y_1(s) - 4Y_2(s) = 3.
\]
Solve for \( Y_2(s) \):

\[
Y_2(s) = \frac{(s + 1) \frac{3s + 22}{s^2 + 3s - 10} - 3}{4}.
\]

### Step 4: Inverse Laplace Transform
Find \( y_1(t) \) and \( y_2(t) \) by taking the inverse Laplace transform of \( Y_1(s) \) and \( Y_2(s) \).

I will now proceed to solve these steps in detail for a full solution.The solutions for \( Y_1(s) \) and \( Y_2(s) \) in the Laplace domain are:

\[
Y_1(s) = \frac{3s + 22}{s^2 + 3s - 10}, \quad Y_2(s) = \frac{4s + 13}{s^2 + 3s - 10}.
\]

### Step 4: Inverse Laplace Transform
Now we

Using the Laplace transform and showing the details of your work, solve the IVP: y1' = -y1 + 4y2, y1(0) = 3; y2' = 3y1 - 2y2, y2(0) = 4

This page provides a detailed step-by-step solution to a system of linear differential equations using the Laplace transform method. The problem involves solving two first-order equations with given initial conditions.

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