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Math Help - Check IVP

  1. #1
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    Check IVP

    I was wondering if someone could check my diff eq I had to solve with an initial condition:

    dy/dt + 2y = 1, y(0) = 5/2

    MY WORK:

    dy/dt = 1 - 2y => dy = (1 - 2y) dt => dt = 1/(1 - 2y) dy

    Take integral of both sides:

    t = -1/2 * ln|2y - 1| + C

    Using the initial condition:

    0 = -1/2 * ln|2*0 - 1| + C

    0 = 0 + C

    Thus C = 0.

    And therefore the solution is t = -1/2 * ln|2y - 1|.

    Anyone see any mistakes?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Ideasman View Post
    I was wondering if someone could check my diff eq I had to solve with an initial condition:

    dy/dt + 2y = 1, y(0) = 5/2

    MY WORK:

    dy/dt = 1 - 2y => dy = (1 - 2y) dt => dt = 1/(1 - 2y) dy

    Take integral of both sides:

    t = -1/2 * ln|2y - 1| + C

    Using the initial condition:

    0 = -1/2 * ln|2*0 - 1| + C

    0 = 0 + C

    Thus C = 0.

    And therefore the solution is t = -1/2 * ln|2y - 1|.

    Anyone see any mistakes?
    you should have solved for y. and you plugged in the initial conditions wrong. (y = 5/2 when t = 0, you plugged in y = t = 0)

    this is a regular first-order inhomogeneous equation. use the method of integrating factor
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  3. #3
    MHF Contributor red_dog's Avatar
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    The mistake is when you found C.
    Put t=0 and \displaystyle y=\frac{5}{2}:
    \displaystyle 0=-\frac{1}{2}\ln 4+C=-\ln 2+C\Rightarrow C=\ln 2
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  4. #4
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    Quote Originally Posted by Jhevon View Post
    you should have solved for y. and you plugged in the initial conditions wrong. (y = 5/2 when t = 0, you plugged in y = t = 0)

    this is a regular first-order inhomogeneous equation. use the method of integrating factor
    Oh, y is a function of t! And I see that I messed up plugging in the values! Let me try re-work it. I'm not quite sure what the method of integrating factor is.
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Ideasman View Post
    I'm not quite sure what the method of integrating factor is.
    see the first post here. make sure it's on page 3
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  6. #6
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    Quote Originally Posted by Jhevon View Post
    you should have solved for y. and you plugged in the initial conditions wrong. (y = 5/2 when t = 0, you plugged in y = t = 0)

    this is a regular first-order inhomogeneous equation. use the method of integrating factor
    Hmm. So I have dy/dt = 1-2y.. then 1/(1-2y) dy = dt.. so the integrals should both stay the same.

    That is:

    t = -1/2 * ln|2y - 1| + C

    So do I solve for y here first? Does it even matter? Can't I just plug the values in now?
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  7. #7
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    Quote Originally Posted by Jhevon View Post
    see the first post here. make sure it's on page 3
    Oh, I think we did this in class! It's used for homogeneous terms of the same degree right?

    Is this where you make a substitution: y = ut or t = vy, where u and v are the new dep. variables?
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Ideasman View Post
    Hmm. So I have dy/dt = 1-2y.. then 1/(1-2y) dy = dt.. so the integrals should both stay the same.

    That is:

    t = -1/2 * ln|2y - 1| + C

    So do I solve for y here first? Does it even matter? Can't I just plug the values in now?
    you can plug in the values and solve for C. it's fine. it's just not the way i would do it. and you may be required to give the answer in the form y(t) = something. if that's the case, this method causes too much trouble for you
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  9. #9
    is up to his old tricks again! Jhevon's Avatar
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    post your solution when finished
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  10. #10
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    Quote Originally Posted by Jhevon View Post
    post your solution when finished
    Thanks for your help Jhevon.

    Luckily for me, my prof. doesn't care if its in the form y(t) = something.

    So here goes my solution:

    dy/dt + 2y = 1

    dy/dt = 1 - 2y => dy = (1 - 2y)dt => dt = 1/(1 - 2y) dy

    Take integral of both sides:

    t = -1/2 * ln|2y - 1| + C

    Using the intial condition:

    0 = -1/2 * ln|2*(5/2) - 1| + C

    0 = -ln(2) + C

    => C = ln(2)

    And thus, the solution is:

    t = -1/2 * ln|2y - 1| + ln(2)
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  11. #11
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    You gotta find y in terms of t

    The general solution is y=\frac12+ke^{-2t}
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  12. #12
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Ideasman View Post
    Thanks for your help Jhevon.

    Luckily for me, my prof. doesn't care if its in the form y(t) = something.

    So here goes my solution:

    dy/dt + 2y = 1

    dy/dt = 1 - 2y => dy = (1 - 2y)dt => dt = 1/(1 - 2y) dy

    Take integral of both sides:

    t = -1/2 * ln|2y - 1| + C

    Using the intial condition:

    0 = -1/2 * ln|2*(5/2) - 1| + C

    0 = -ln(2) + C

    => C = ln(2)

    And thus, the solution is:

    t = -1/2 * ln|2y - 1| + ln(2)
    that's fine. good job.

    i got Krizalid's solution though (with k = 2).

    Quote Originally Posted by Krizalid View Post
    You gotta find y in terms of t

    The general solution is y=\frac12+ke^{-2t}
    we can probably turn them into the same thing with some algebraic manipulation, but if your professor doesn't care, why bother
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  13. #13
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    Quote Originally Posted by Jhevon View Post
    that's fine. good job.

    i got Krizalid's solution though (with k = 2).

    we can probably turn them into the same thing with some algebraic manipulation, but if your professor doesn't care, why bother
    So you're saying if I DID solve for y in terms of t, I would get  y=\frac12+ke^{-2t} and with the initial condition, k = 2? So what we have are equivalent?
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  14. #14
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Ideasman View Post
    So you're saying if I DID solve for y in terms of t, I would get  y=\frac12+ke^{-2t} and with the initial condition, k = 2? So what we have are equivalent?
    yes, they are equivalent. I'll show you

    t = - \frac {1}{2} \ln |2y - 1| + \ln 2

    \Rightarrow -2t = \ln |2y - 1| - 2 \ln 2

    \Rightarrow -2t = \ln |2y - 1| - \ln 4

    \Rightarrow -2t = \ln \left| \frac {2y - 1}{4} \right|

    \Rightarrow e^{-2t} = \frac {2y - 1}{4}

    \Rightarrow y = \frac {1}{2} + 2 e^{-2t}

    and, of course, we could go the other way, from my solution to yours
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  15. #15
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    Actually, I solved this one as a linear ODE, it's faster.

    When Ideasman look at y'+P(x)y=Q(x), he will see
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