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Solve the initial value problem?

  1. Jul 11, 2013 #1
    Solve the initial value problem y'=2t(1+y), y(0)=0 by the method of successive approximations.

    I don't know how to do this problem but I think there's integral involved in it. Please help me. Thanks.
     
  2. jcsd
  3. Jul 11, 2013 #2
    Does it have to be done by the method of successive approximations? I mean, it looks fairly simple to just say
    $$\frac{dy}{dt}=2t+2ty \\ \frac{dy}{dt}-2ty=2t \implies \frac{d\mu}{dt}=-2t\mu \implies \mu=c_1e^{-t^2} \\ \frac{d}{dt}\left[c_1e^{-t^2}y\right]=2c_1te^{-t^2} \\ y = 2e^{t^2}\int te^{-t^2}dt \\ y = 2c_2e^{t^2}-1 \\ 0 = 2c_2-1 \implies c_2=\frac{1}{2} \\ y(t)=e^{t^2}-1.$$
     
  4. Jul 11, 2013 #3
    Isn't the equation separable if you wanted to solve it analytically?

    [tex] \frac{dy}{dt}=2t(1+y) \\
    \frac{dy}{1+y}=2t dt \\
    \ln|1+y|=t^2+C \\
    y+1=\pm e^C e^{t^2} [/tex] and using the initial condition gives [tex] y=e^{t^2}-1 [/tex]
     
    Last edited by a moderator: Jul 12, 2013
  5. Jul 11, 2013 #4
    ...But it's more fun to do it the long way! :tongue:

    I need to start seeing when there's an easier way to solve differential equations.

    As a side note, ##\frac{d}{dx}\ln{x}=\frac{1}{x}\implies\int\frac{dx}{x}=\ln{x}+C##. No need for absolute values unless you are adamant about ignoring complex numbers.

    Alright. Let's do it the OP's way. Suppose we have a solution ##y##. Then, for ##\frac{dy}{dt}=f(t,y(t))=2t(1+y)##, we have that ##\displaystyle y=\int\limits_{[t_0,t]}f(x,y(x)) \, dx##. We thus approach by the method of approximations. Define ##\displaystyle y_1=\int\limits_{[t_0,t]}f(x,0) \, dx = \int\limits_{[t_0,t]}2x\, dx = t^2-t_0^2## and ##\displaystyle y_n=\int\limits_{[t_0,t]}f(x,y_{n-1}) \, dx##.

    Is everything clear so far? How might you continue this approach, Success?
     
    Last edited by a moderator: Jul 12, 2013
  6. Jul 12, 2013 #5
    Seeing as it's been almost a day, I will suggest that we set ##t_0=0## and point out that ##\displaystyle y_n = \sum_{k=1}^{n}\left[\frac{t^{2k}}{k!}\right]##.
     
  7. Jul 12, 2013 #6
    Mandel, what's y1 and yn then in this problem?
     
  8. Jul 12, 2013 #7

    micromass

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    Sucess, you have received several hints right now. I would like to hear some kind of effort from you first before we continue.
     
  9. Jul 12, 2013 #8
    y=2 integral of t(1+y)dt from 0 to t. Now I need y1 and y2, I'll try. Give me some minutes.
     
  10. Jul 12, 2013 #9
    y=2 integral of (t+ty)dt from 0 to t. How to integrate this?
     
  11. Jul 12, 2013 #10

    HallsofIvy

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    You don't. y is an unknown function of t. You cannot integrate an unknown function!

    People have been trying to tell you to separate the y and t terms:
    If dy/dt= t(1+ y) then dy/(1+ y)= t dt.
    [tex]\int \frac{dy}{1+ y}= \int t dt[/tex].
     
  12. Jul 12, 2013 #11

    Mark44

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    Start with the simplest method first. In this case, it's easy to see by inspection that the equation is separable. The two factors on the right side were already separate functions of y and t, so multiplying out the two factors made it less obvious that separation could be used.
     
  13. Jul 12, 2013 #12

    D H

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    All of the hints so far are completely off course because they don't address the central issue of using the method of successive approximations. This approach finds a sequence of functions that hopefully converge to the solution. It is very reminiscent of fixed point iteration.

    Start with some guess, call it y0(t), that satisfies the initial conditions. There's an obvious one here: y0(t)=0. Next iterate by solving dyn+1(t)/dt = 2t(1+yn), yn+1(0)=0. The first step is easy: Just solve dy1(t)/dt=2t(1+0)=2t, y1(0)=0. Then keep iterating. You should see a power series arising.
     
  14. Jul 12, 2013 #13
    Thank you guys. The one HS-Scientist did was very understandable.
     
  15. Jul 12, 2013 #14

    D H

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    I suspect you won't get very much credit if you submit his solution. While it is the solution to the initial value problem, it does not use the method of successive approximations.
     
  16. Jul 12, 2013 #15
    But why do we must use a hard method when we can do it using the easier one?
     
  17. Jul 12, 2013 #16

    D H

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    Because sometimes there is no clear-cut analytic solution.
     
  18. Jul 12, 2013 #17

    Mark44

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    Which is something that was nagging me in the back of my mind.

    Aside from what D H mentioned, because that is the method requested in the problem statement (emphasis added).
     
  19. Jul 13, 2013 #18

    D H

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    As an example of how to use the method of successive approximations, consider the ODE y'=y, y(0)=1. Note that this is not the problem posed in the OP. Success, it's up to you to adapt this to the problem at hand.

    What this method does is to build a sequence of functions y0(t), y1(t), y2(t), etc., each step being a bit closer to the solution than the previous. The relation between successive approximations is yn'(t) = yn-1(t) such that yn(0)=1. The starting point is a constant function that satisfies the initial condition. In this example, y(0)=1, so y0(t) = 1. This gives the conditions for the next approximation: y1'(t)=1, y1(0)=1. The solution to this IVP is y1(t)=1+t. The next iteration yields y2(t)=1+t+t2/2, the next y3(t)=1+t+t2/2+t3/6.

    Notice how each step builds toward toward the infinite series [itex]y(t)=\sum_{n=0}^{\infty} \frac{t^n}{n!}[/itex]
     
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