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Math Help - Verifying the solution of a differential equation.

  1. #1
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    Verifying the solution of a differential equation.

    I have the following differential equation >> dx/dt = -1/2 (x)
    it says i need to verify that x(t)=Ce^-t/2

    i have done:

    rearrange >> 1/x dx = -1/2 dt
    integrate >> ln x = -t/2 + C
    take log to simplify >> ln x = ln e^-t/2 + ln C
    >> ln x = ln (Ce^-t/2)
    which finally gives >> x = Ce^-t/2

    is this correct or am i doing it wrong??

    It then goes on to say find the particular solution that satisfies x(0) = 2 and sketch on a direction field...so:

    input the condition >> 2 = Ce^-0/2
    simplify >> 2 = Ce^0
    >> 2 = C(1)
    therefore >> 2 = C
    and the particular solution is >> x = 2e^-t/2

    again is this correct? And how would i go about sketching this?

    Thanks
    Last edited by mr fantastic; August 8th 2011 at 10:37 PM. Reason: Title.
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  2. #2
    Grand Panjandrum
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    Re: General solution of differential equation

    Quote Originally Posted by NFS1 View Post
    I have the following differential equation >> dx/dt = -1/2 (x)
    it says i need to verify that x(t)=Ce^-t/2
    With a question like this, where you are asked to verify that a proposed solution is in fact a solution you just need to show that the given solution satisfies the ODE.

    So differentiate to get:

    x'(t)=-\frac{1}{2}Ce^{-t/2}

    which we see is only a solution if C=1 (that is the arbitary constant is in the wrong place)

    The correct solution is:  x(t)=C+e^{-t/2}

    CB
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  3. #3
    MHF Contributor Siron's Avatar
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    Re: General solution of differential equation

    @CB:
    Can you explain why wolphram alpha comes to the solution:
    dx/dt=-x/2 - Wolfram|Alpha

    Is that similar, or?
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  4. #4
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    Re: General solution of differential equation

    Quote Originally Posted by CaptainBlack View Post
    With a question like this, where you are asked to verify that a proposed solution is in fact a solution you just need to show that the given solution satisfies the ODE.

    So differentiate to get:

    x'(t)=-\frac{1}{2}Ce^{-t/2}

    which we see is only a solution if C=1 (that is the arbitary constant is in the wrong place)

    The correct solution is:  x(t)=C+e^{-t/2}

    CB
    The way i have done is incorrect then? I have looked through my uni notes and thought the way i have done it would be correct. Also when i have substituted the intial condition of x(0)=2 is that correct?
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  5. #5
    MHF Contributor Siron's Avatar
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    Re: General solution of differential equation

    Quote Originally Posted by NFS1 View Post
    rearrange >> 1/x dx = -1/2 dt
    integrate >> ln x = -t/2 + C
    take log to simplify >> ln x = ln e^-t/2 + ln C
    I would change C into (for example) K, because in this step you show that \ln(C)=C \Leftrightarrow e^{C}=C, which is not true, sou I would say:
    \ln(x)=\ln(e^{\frac{-t}{2}})+\ln(K) where K is a new constant, in fact: K=e^{C}.

    Or you just can write:
    \ln(x)=\ln(e^{\frac{-t}{2}})+\ln(e^{C})
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  6. #6
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    Re: General solution of differential equation

    Quote Originally Posted by NFS1 View Post
    I have the following differential equation >> dx/dt = -1/2 (x)
    it says i need to verify that x(t)=Ce^-t/2

    i have done:

    rearrange >> 1/x dx = -1/2 dt
    integrate >> ln x = -t/2 + C
    take log to simplify >> ln x = ln e^-t/2 + ln C
    >> ln x = ln (Ce^-t/2)
    which finally gives >> x = Ce^-t/2

    is this correct or am i doing it wrong??

    It then goes on to say find the particular solution that satisfies x(0) = 2 and sketch on a direction field...so:

    input the condition >> 2 = Ce^-0/2
    simplify >> 2 = Ce^0
    >> 2 = C(1)
    therefore >> 2 = C
    and the particular solution is >> x = 2e^-t/2

    again is this correct? And how would i go about sketching this?

    Thanks
    First of all, don't forget that the integral of 1/x is ln|x|, the modulus is important.

    Second, why not just exponentiate both sides to simplify?

    \displaystyle \begin{align*} \ln{|x|} &= -\frac{t}{2} + c \\ |x| &= e^{-\frac{t}{2} + c} \\ |x| &= e^ce^{-\frac{t}{2}} \\ x &= \pm e^ce^{-\frac{t}{2}} \\ x &= Ce^{-\frac{t}{2}}\textrm{ where }C = \pm e^c\end{align*}
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  7. #7
    Grand Panjandrum
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    Re: General solution of differential equation

    Quote Originally Posted by NFS1 View Post
    The way i have done is incorrect then? I have looked through my uni notes and thought the way i have done it would be correct. Also when i have substituted the intial condition of x(0)=2 is that correct?
    It is a question of technique, if asked to veryfy something you can just check that it is a solution, you do not need to solve the problem from scratch.

    CB
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  8. #8
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    Re: General solution of differential equation

    Quote Originally Posted by Prove It View Post
    First of all, don't forget that the integral of 1/x is ln|x|, the modulus is important.

    Second, why not just exponentiate both sides to simplify?

    \displaystyle \begin{align*} \ln{|x|} &= -\frac{t}{2} + c \\ |x| &= e^{-\frac{t}{2} + c} \\ |x| &= e^ce^{-\frac{t}{2}} \\ x &= \pm e^ce^{-\frac{t}{2}} \\ x &= Ce^{-\frac{t}{2}}\textrm{ where }C = \pm e^c\end{align*}
    Thankyou mate. Makes alot more sense. So am i writing in say the particular solution is x=2e^-t/2 if the initial condition is x(0)=2??
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  9. #9
    MHF Contributor Siron's Avatar
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    Re: General solution of differential equation

    I thinkt that's correct.
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