Question 1184044: Find a solution to the initial value problem
y′+sin(t)y=g(t), y(0)=7,
that is continuous on the interval [0,2π] where
g(t)={sin(t), if 0≤t≤π
-----{ −sin(t),if π less than"t" less than or equal to 2pi
y(t)={ My answer is 6e^(-2)+1, but is wrong,if 0≤t≤π.
---{ My answer is (6e^(-2)+2e)e^(cost)-1, but wrong, if π less than"t" less than or equal to 2pi
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! 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.
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