Solve the initial value problem $$ \frac{d y}{d t}-y=2 e^{t}+4 e^{3 t}, \quad y(0)=8 $$ $$ y(t)= $$

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We are given the differential equation
$$
\frac{dy}{dt} - y = 2e^{t} + 4e^{3t}, \quad y(0) = 8.
$$
This is a first-order linear differential equation. We solve it using the integrating factor method.

**Step 1. Write the equation in standard form**

The standard form of a first-order linear ODE is
$$
\frac{dy}{dt} + p(t)y = q(t).
$$
For our equation, we have
$$
\frac{dy}{dt} - y = 2e^{t} + 4e^{3t},
$$
so
$$
p(t) = -1 \quad \text{and} \quad q(t) = 2e^{t} + 4e^{3t}.
$$

**Step 2. Determine the integrating factor**

The integrating factor $$ \mu(t) $$ is given by
$$
\mu(t) = e^{\int p(t)\,dt} = e^{\int -1\,dt} = e^{-t}.
$$

**Step 3. Multiply the differential equation by the integrating factor**

Multiply both sides of the differential equation by $$ e^{-t} $$:
$$
e^{-t}\frac{dy}{dt} - e^{-t}y = 2e^{t}\cdot e^{-t} + 4e^{3t}\cdot e^{-t}.
$$
This simplifies to:
$$
e^{-t}\frac{dy}{dt} - e^{-t}y = 2 + 4e^{2t}.
$$

Notice that the left-hand side is the derivative of $$ y e^{-t} $$. In fact,
$$
\frac{d}{dt}\bigl(y e^{-t}\bigr) = e^{-t}\frac{dy}{dt} - e^{-t}y.
$$

Thus, we have:
$$
\frac{d}{dt}\bigl(y e^{-t}\bigr) = 2 + 4e^{2t}.
$$

**Step 4. Integrate both sides**

Integrate with respect to $$ t $$:
$$
\int \frac{d}{dt}\bigl(y e^{-t}\bigr)\,dt = \int \left(2 + 4e^{2t}\right) \, dt.
$$
The left-hand side integrates to:
$$
y e^{-t}.
$$
For the right-hand side, we integrate term by term:
$$
\int 2 \, dt = 2t,
$$
and
$$
\int 4e^{2t}\, dt = 4 \cdot \frac{1}{2} e^{2t} = 2e^{2t}.
$$
Thus, we obtain:
$$
y e^{-t} = 2t + 2e^{2t} + C,
$$
where $$ C $$ is the constant of integration.

**Step 5. Solve for $$ y(t) $$**

Multiply both sides by $$ e^{t} $$ to solve for $$ y(t) $$:
$$
y(t) = e^{t}\bigl(2t + 2e^{2t} + C\bigr) = 2te^{t} + 2e^{3t} + C e^{t}.
$$

**Step 6. Apply the initial condition**

We use the initial condition $$ y(0) = 8 $$:
$$
y(0) = 2\cdot 0\cdot e^{0} + 2e^{0} + C e^{0} = 0 + 2 + C = 8.
$$
Thus,
$$
C = 8 - 2 = 6.
$$

**Final Answer**

The solution to the initial value problem is:
$$
y(t) = 2te^{t} + 2e^{3t} + 6e^{t}.
$$
$$ y(t) = 2te^{t} + 2e^{3t} + 6e^{t} $$

Solve the initial value problem \( \frac{d y}{d t}-y=2 e^{t}+4 e^{3 t}, \quad y(0)=8 \) \( y(t)= \)

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