- #1

thedude36

- 30

- 0

y' = 1-t+y , y(t

_{0})=y_{0}show that the exact solution is

y=[itex]\phi[/itex](t)=(y

_{0}-t_{0})e^{t-t0}+twe've only spoken of approximations in class and I've just been kind of guessing as to how i should go about it so far. I've tried to look up the term exact solutions but haven't found anything of much use.

## Homework Equations

integrating factor: if dy/dt+ay=g(t), then μ(t) is such that dμ(t)/dt=aμ(t). Multiply both sides of the equation dy/dt+ay=g(t) by μ(t) to obtain μ(t)dy/dt+ayμ(t)=μ(t)g(t).

## The Attempt at a Solution

rearranging the given formula, i was able to get

dy/dt-y=1-t

where a = -1. thus, dμ(t)/dt=-μ(t) making μ(t)=e

^{-t}. multiplying bothsides give

e

^{-t}dy/dt-e^{-t}y=e^{-t}-te^{-t}the left side can be obtained by the power-rule if the initial function was d(ye

^{-t})/dt so we replace the leftside with this

d(ye

^{-t})/dt=e^{-t}-te^{-t}integrating bothsides and then solving for y gives

y=-2-t+ce

^{t}however, i don't think this is the correct method. What should i be doing instead of this?