Solve the initial value problem x’=3x+y-e^3t x(0)=2 y’=x+3y y(0)=-2

We are given a system of differential equations with initial conditions:

\[
x' = 3x + y - e^{3t}, \quad x(0) = 2
\]
\[
y' = x + 3y, \quad y(0) = -2
\]

### Step 1: Solve the second equation \( y' = x + 3y \)

We first assume that \(x(t)\) is known and solve for \(y(t)\). The equation \(y' = x + 3y\) is a first-order linear differential equation in \(y\). We can use the method of integrating factors to solve it.

The equation can be rewritten as:
\[
y' - 3y = x
\]

The integrating factor is:
\[
\mu(t) = e^{-3t}
\]

Multiplying through the equation by the integrating factor:
\[
e^{-3t}y' - 3e^{-3t}y = e^{-3t}x
\]

This simplifies to:
\[
\frac{d}{dt}\left( e^{-3t}y \right) = e^{-3t}x
\]

Now integrate both sides:
\[
e^{-3t}y = \int e^{-3t}x \, dt
\]

At this point, we need \(x(t)\), so we now focus on solving for \(x(t)\).

---

### Step 2: Solve the first equation \( x' = 3x + y - e^{3t} \)

We treat this as a system and try to solve for \(x(t)\) and \(y(t)\) together. This can be approached using methods for solving linear systems, like undetermined coefficients or variation of parameters.

Let me proceed by solving the system step by step.The general solutions to the system of differential equations are:

\[
x(t) = -C_1 e^{2t} + C_2 e^{4t}
\]
\[
y(t) = C_1 e^{2t} + C_2 e^{4t} + e^{3t}
\]

### Step 3: Apply the initial conditions

We are given the initial conditions:
\[
x(0) = 2, \quad y(0) = -2
\]

Substitute \(t = 0\) into the general solutions:

For \(x(0) = 2\):
\[
x(0) = -C_1 e^{0} + C_2 e^{0} = -C_1 + C_2 = 2
\]
Thus, we have the equation:
\[
-C_1 + C_2 = 2 \quad \text{(Equation 1)}
\]

For \(y(0) = -2\):
\[
y(0) = C_1 e^{0} + C_2 e^{0} + e^{3 \cdot 0} = C_1 + C_2 + 1 = -2
\]
Thus, we have the equation:
\[
C_1 + C_2 + 1 = -2 \quad \Rightarrow \quad C_1 + C_2 = -3 \quad \text{(Equation 2)}
\]

### Step 4: Solve for the constants \(C_1\) and \(C_2\)

Now we solve the system of equations:
\[
\begin{aligned}
-C_1 + C_2 &= 2 \\
C_1 + C_2 &= -3
\end{aligned}
\]

Solving this system will give us the values of \(C_1\) and \(C_2\).The values of the constants are:

\[
C_1 = -\frac{5}{2}, \quad C_2 = -\frac{1}{2}
\]

### Step 5: Substitute the constants into the general solution

Substitute \(C_1\) and \(C_2\) into the general solutions for \(x(t)\) and \(y(t)\):

\[
x(t) = -\left(-\frac{5}{2}\right) e^{2t} + \left(-\frac{1}{2}\right) e^{4t} = \frac{5}{2} e^{2t} - \frac{1}{2} e^{4t}
\]

\[
y(t) = \left(-\frac{5}{2}\right) e^{2t} + \left(-\frac{1}{2}\right) e^{4t} + e^{3t} = -\frac{5}{2} e^{2t} - \frac{1}{2} e^{4t} + e^{3t}
\]

Thus, the solutions to the initial value problem are:

\[
x(t) = \frac{5}{2} e^{2t} - \frac{1}{2} e^{4t}
\]

\[
y(t) = -\frac{5}{2} e^{2t} - \frac{1}{2} e^{4t} + e^{3t}
\]

Would you like further details or have any questions?

### Related questions:
1. How does the method of integrating factors work for first-order linear differential equations?
2. What are the conditions under which a system of linear differential equations has unique solutions?
3. Can you solve this system using matrix methods (e.g., eigenvalue decomposition)?
4. How would you handle non-homogeneous terms in a system of differential equations differently?
5. What is the general approach for solving higher-order systems of differential equations?

**Tip:** Always check the consistency of initial conditions after solving a system of differential equations to verify your solution.

Solve the initial value problem x’=3x+y-e^3t, x(0)=2, y’=x+3y, y(0)=-2.

Explore a step-by-step solution to a system of differential equations with initial conditions. This problem covers topics like integrating factors, solving systems of equations, and exponential functions. Ideal for advanced university students.

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