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

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: \[ y(t) = 2te^{t} + 2e^{3t} + 6e^{t}. \]