Question 1184044
Here's how to solve the initial value problem:

**1. Solve for 0 ≤ t ≤ π:**

The equation is y' + sin(t)y = sin(t). This is a first-order linear differential equation. We can find the integrating factor:

Integrating factor = e^(∫sin(t)dt) = e^(-cos(t))

Multiply the equation by the integrating factor:

e^(-cos(t))y' + sin(t)e^(-cos(t))y = sin(t)e^(-cos(t))

Notice that the left side is the derivative of (e^(-cos(t))y):

d/dt (e^(-cos(t))y) = sin(t)e^(-cos(t))

Integrate both sides with respect to t:

∫ d/dt (e^(-cos(t))y) dt = ∫ sin(t)e^(-cos(t)) dt

e^(-cos(t))y = e^(-cos(t)) + C₁

y(t) = 1 + C₁e^(cos(t))

Using the initial condition y(0) = 7:

7 = 1 + C₁e^(cos(0))
7 = 1 + C₁e
C₁ = 6/e

So, for 0 ≤ t ≤ π:

y(t) = 1 + (6/e)e^(cos(t)) = 1 + 6e^(cos(t)-1)

**2. Solve for π < t ≤ 2π:**

The equation is y' + sin(t)y = -sin(t).  The integrating factor is the same: e^(-cos(t))

Multiply the equation by the integrating factor:

e^(-cos(t))y' + sin(t)e^(-cos(t))y = -sin(t)e^(-cos(t))

d/dt (e^(-cos(t))y) = -sin(t)e^(-cos(t))

Integrate both sides:

e^(-cos(t))y = e^(-cos(t)) + C₂

y(t) = 1 + C₂e^(cos(t))

To find C₂, we need to use the continuity of y at t = π.

At t = π, the solutions from both intervals must be equal:

1 + 6e^(cos(π)-1) = 1 + C₂e^(cos(π))
1 + 6e^(-2) = 1 + C₂e^(-1)
6e^(-2) = C₂e^(-1)
C₂ = 6/e

So, for π < t ≤ 2π:

y(t) = 1 + (6/e)e^(cos(t)) = 1 + 6e^(cos(t)-1)

**Final Solution:**

y(t) = 1 + 6e^(cos(t) - 1),  0 ≤ t ≤ 2π

Your original answer was close, but you made a mistake in the calculation of the constant C for the second interval. The correct solution shows that the function has the same form in both intervals, ensuring continuity.