Exact solutions and the convergence of eulers method

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SUMMARY

The discussion centers on solving the initial value problem defined by the differential equation y' = 1 - t + y, with the initial condition y(t0) = y0. The exact solution is confirmed to be y = φ(t) = (y0 - t0)e^t - t0 + t. The user initially struggled with the integration process but was guided to use the integrating factor method, specifically μ(t) = e^-t, to simplify the equation. A mistake in the integration step was identified and corrected, leading to a successful resolution of the problem.

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im having trouble with this question - http://i.imgur.com/Ars4J1b.png - more specifically with part a, as i have a good idea how to go about b. given the initial value problem

y' = 1-t+y , y(t0)=y0

show that the exact solution is

y=[itex]\phi[/itex](t)=(y0-t0)et-t0+t​

we'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-tdy/dt-e-ty=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+cet

however, i don't think this is the correct method. What should i be doing instead of this?
 
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Your method is fine. You just have a mistake calculating your last line. Re-check that -2-t.
 
LCKurtz said:
Your method is fine. You just have a mistake calculating your last line. Re-check that -2-t.

Thanks! I am not sure how i was integrating that wrong, but a quick check on a calculator quickly resolved the issue. I appreciate it!
 

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