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Find a Solution to the Initial Value Problem That is Continuous on the Interval

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Find a solution to the initial value problem

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Homework Statement


Find a solution to the initial value problem
53472edda767ff31eafec295f8d4491.png

that is continuous on the interval
1d5856a507737b03db7c79778851561.png
where
21564751f7c18c60a1a118c765afae1.png

Homework Equations


I know the equations, but don't want to type them out.

The Attempt at a Solution

I got the first part of this question. The part where g(t) = sin(t)

I can not figure it out when g(t)=-sin(t)
The answer I have come up with for that part is y(t)= -1+(8*e^(cos(t))/e)
This is using y(0)=y(2pi)

Any reason I can't do this? There must be some reason because I'm not getting the right answer.

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Answers and Replies

Okay so I'll assume you've actually solved y' + sin(t)y = sin(t) for 0 ≤ t ≤ π.

I got y(t) = cecos(t) + 1. As the general solution.

As for solving the second one, I got y(t) = cecos(t) - 1 for π ≤ t ≤ 2π.

I'm not sure how you're getting what you got?

I have solved for y' + sin(t)y = sin(t) for 0 ≤ t ≤ π.

Since we are given y(0)=7 we are not allowed to put C in our answer. SO my First answer was y=1+6/e*e^cos(t)

but since we are not given something for the other interval I don't know how to solve for C. I thought I could use y(0) = 7 since 2pi and 0 are equal for sin and cos, but it is not right.

I have solved for y' + sin(t)y = sin(t) for 0 ≤ t ≤ π.

Since we are given y(0)=7 we are not allowed to put C in our answer. SO my First answer was y=1+6/e*e^cos(t)

but since we are not given something for the other interval I don't know how to solve for C. I thought I could use y(0) = 7 since 2pi and 0 are equal for sin and cos, but it is not right.


Your first answer is not correct, you may want to check it again, I'll assume it's an arithmetic error.

Also, you have the right idea evaluating the second one at 2π, once again though it's probably just arithmetic.

Your first answer is not correct, you may want to check it again, I'll assume it's an arithmetic error.

Also, you have the right idea evaluating the second one at 2π, once again though it's probably just arithmetic.


I'm not sure how my first is wrong? This is for online homework and I put it in and it says it is correct.
I have solved for y' + sin(t)y = sin(t) for 0 ≤ t ≤ π.

Since we are given y(0)=7 we are not allowed to put C in our answer. SO my First answer was y=1+6/e*e^cos(t)

but since we are not given something for the other interval I don't know how to solve for C. I thought I could use y(0) = 7 since 2pi and 0 are equal for sin and cos, but it is not right.


Yes, y=1+(6/e)*e^cos(t) is correct for the first part. Use that to find y(pi) and use it as an initial value for the second. And no, you can't assume y(0)=y(2pi). It's true for the individual parts. But it's not necessarily true if you are splicing them together and you cross the point where you joined them.
I tried a quick and dirty Excel brute force solution with just 10 points in each interval.

Wonder how off I am?

Mib89aV.png

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