To solve this initial value problem (IVP), let's analyze the given differential equation and apply the method for solving first-order linear differential equations.

The problem is: $y' + y = e^{11t}, \quad y(0) = -1$

Step 1: Identify the Integrating Factor

This is a linear differential equation of the form: $y' + p(t) y = g(t)$ where $p(t) = 1$ and $g(t) = e^{11t}$.

To solve, we need an integrating factor $\mu(t)$, defined as: $\mu(t) = e^{\int p(t) \, dt} = e^{\int 1 \, dt} = e^t$

Step 2: Multiply through by the Integrating Factor

Multiply both sides of the equation by $e^t$: $e^t y' + e^t y = e^t \cdot e^{11t}$ which simplifies to: $e^t y' + e^t y = e^{12t}$

Step 3: Rewrite as the Derivative of a Product

The left side of the equation is now the derivative of $e^t y$: $\frac{d}{dt}(e^t y) = e^{12t}$

Step 4: Integrate Both Sides

Integrate both sides with respect to $t$: $\int \frac{d}{dt}(e^t y) \, dt = \int e^{12t} \, dt$ This gives: $e^t y = \frac{e^{12t}}{12} + C$

Step 5: Solve for $y$

Now, divide by $e^t$ to isolate $y$: $y = \frac{e^{12t}}{12 e^t} + \frac{C}{e^t} = \frac{e^{11t}}{12} + Ce^{-t}$

Step 6: Apply the Initial Condition

Use the initial condition $y(0) = -1$ to find $C$: $y(0) = \frac{e^{11 \cdot 0}}{12} + C \cdot e^{-0} = \frac{1}{12} + C = -1$ Solving for $C$: $C = -1 - \frac{1}{12} = -\frac{12 + 1}{12} = -\frac{13}{12}$

Final Solution

Substitute $C = -\frac{13}{12}$ back into the expression for $y$: $y = \frac{e^{11t}}{12} - \frac{13}{12} e^{-t}$

So, the solution to the initial value problem is: $y = \frac{e^{11t}}{12} - \frac{13}{12} e^{-t}$

Would you like further details on any part of this solution?

Related Questions

1. How can I determine the integrating factor for a general first-order linear differential equation?
2. What is the significance of the initial condition in solving differential equations?
3. How do I interpret the solution in terms of exponential growth and decay?
4. What happens to the solution as $t \to \infty$?
5. How do initial conditions affect the constants in solutions to differential equations?

Tip

When solving differential equations, always verify the solution by substituting it back into the original equation.