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The given equation is:

\[
\dot{y} + 2y = -t e^{-2t}, \quad y(1) = 0
\]

This is a first-order linear differential equation with the initial condition \( y(1) = 0 \).

### Step 1: Solve the homogeneous equation

First, solve the homogeneous equation:

\[
\dot{y} + 2y = 0
\]

The characteristic equation for this is:

\[
r + 2 = 0 \quad \Rightarrow \quad r = -2
\]

So, the solution to the homogeneous equation is:

\[
y_h(t) = C e^{-2t}
\]

### Step 2: Solve the non-homogeneous equation

Now, for the non-homogeneous part, we use an **integrating factor** method.

The integrating factor \( \mu(t) \) is given by:

\[
\mu(t) = e^{\int 2 \, dt} = e^{2t}
\]

Multiplying the entire differential equation by this integrating factor:

\[
e^{2t} \dot{y} + 2 e^{2t} y = -t
\]

This simplifies to:

\[
\frac{d}{dt} \left( e^{2t} y \right) = -t
\]

### Step 3: Integrate both sides

Now integrate both sides with respect to \( t \):

\[
e^{2t} y = \int -t \, dt = -\frac{t^2}{2} + C
\]

So:

\[
y(t) = e^{-2t} \left( -\frac{t^2}{2} + C \right)
\]

### Step 4: Apply the initial condition

Use the initial condition \( y(1) = 0 \) to find \( C \):

\[
0 = e^{-2(1)} \left( -\frac{1^2}{2} + C \right)
\]

\[
0 = e^{-2} \left( -\frac{1}{2} + C \right)
\]

\[
C = \frac{1}{2}
\]

### Final solution:

The general solution is:

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

This is the solution to the differential equation with the given initial condition.

Would you like further details or have any questions? Here are 5 related questions to consider:

1. How can the integrating factor method be applied to second-order differential equations?
2. What is the importance of homogeneous solutions in differential equations?
3. Could this problem be solved using another method, such as undetermined coefficients?
4. How would the solution change if the initial condition was different, say \( y(0) = 1 \)?
5. What are the common applications of first-order linear differential equations?

**Tip:** Always verify your final solution by substituting it back into the original equation to ensure it satisfies both the differential equation and the initial condition.

Solve the differential equation y' + 2y = -te^(-2t), y(1) = 0.

Learn how to solve the first-order linear differential equation y' + 2y = -te^(-2t) with the initial condition y(1) = 0. This problem covers key topics like the integrating factor method and the solving process for non-homogeneous differential equations. Suitable for college-level calculus students.

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