Exact solutions and the convergence of eulers method

  • Thread starter Thread starter thedude36
  • Start date Start date
  • Tags Tags
    Convergence Method
thedude36
Messages
30
Reaction score
0
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=\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?
 
Physics news on Phys.org
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!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top