thedude36
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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
show that the exact solution is
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.
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).
rearranging the given formula, i was able to get
where a = -1. thus, dμ(t)/dt=-μ(t) making μ(t)=e-t. multiplying bothsides give
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
integrating bothsides and then solving for y gives
however, i don't think this is the correct method. What should i be doing instead of this?
y' = 1-t+y , y(t0)=y0
show that the exact solution is
y=\phi(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?