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

[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.

Homework Equations

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...

 

[Dick]

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.