# General solution of initial value problem -dont understand problem is asking me?

• dwilmer
In summary, the problem is asking for a value of y-sub-0 that will make the solution remain finite as t approaches infinity. The general solution is found, but the correct value for y-sub-0 is needed to make the solution work. The book answer suggests y-sub-0 = -5/2, but it is unclear how or why. The integrating factor should be e^(-t) and the overall concept is to arrange the initial conditions to set c=0, so that the solution remains finite as t approaches infinity.
dwilmer
General solution of initial value problem --dont understand problem is asking me??

## Homework Statement

Find a value for y-sub-0 for which the solution of the initial value problem:
y' - y = 1+ 3sin t y(0) - y-sub-0
remains finite as t approaches infinity.

(i called it "y-sub-0" , just because i can't do subscript)

book answer says y-sub-0 = -5/2 but i don't understand how or why they have that.

## The Attempt at a Solution

first of all, i got the general solution (but then i am stuck)

y' - y = 1+ 3sin t

get integrating factor...
u(t) = e^t

(e^t)y = integ: (1+ 3sin t)e^t

expand RHS...

(e^t)y = integ: e^t + integ (e^t)(3sin t)

integrate RHS using integ by parts on the last term of RHS (this is where i suspect i am messing up??)
(e^t)y = e^t + 3/4e^t(sin t + cos t) + e^t + c

divide both sides by e^t to isolate y, and cancel where applicable

y = e^t + 3/4(sin t - cos t) + 1 + c/e^t

then, apply initial condition , y(0) = ysub0

y = 1/4 + c

BUT --this is where i am confused. because the question asks what value of y will remain finite as t approaches infinity? But they just said that t = 0 in the initial conditions ( y(0) = y-sub-0 )

so i don't understand what they are asking me.. can you please rephrase what value i am looking for?

book answer says y = -5/2, but i don't see how or why that is the answer...

For one thing, the integrating factor should be e^(-t), shouldn't it? As for the overall concept, if you find the correct solution, it should have a term like c*e^(t). e^(t) goes to infinity, so you want to arrange your initial conditions so that you can set c=0.

## 1. What is a general solution in the context of initial value problems?

A general solution is a solution that contains all possible solutions to an initial value problem. It includes a constant that can take on different values for different specific solutions.

## 2. How can a general solution be obtained from an initial value problem?

A general solution can be obtained by solving the differential equation that represents the initial value problem and then adding a constant to the solution.

## 3. What is the significance of the initial value in an initial value problem?

The initial value represents the starting point for the solution of a differential equation. It is used to determine a specific solution from the general solution.

## 4. What is the difference between a general solution and a specific solution?

A general solution contains all possible solutions to an initial value problem and includes a constant, while a specific solution is obtained by substituting a specific value for the constant in the general solution.

## 5. Can a general solution be used to find a specific solution to an initial value problem?

Yes, a specific solution can be obtained from the general solution by substituting a specific value for the constant. This allows for a more efficient method of finding solutions to initial value problems.

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