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Math Help - Second Derivative with only "y" variable

  1. #1
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    Second Derivative with only "y" variable

    How do you take the 2nd derivative for a function without an x on the right side of the equation?

    Ie: y'= 0.05y with y(0)=1000 which would yield y'(0)=50, right?

    How do I find y'' for this equation?

    Thank you,
    Jen
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Jenberl View Post
    How do you take the 2nd derivative for a function without an x on the right side of the equation?

    Ie: y'= 0.05y with y(0)=1000 which would yield y'(0)=50, right?

    How do I find y'' for this equation?

    Thank you,
    Jen

    Hmm...this is what I think you should do.

    I suggest you solve the differential equation \frac{\,dy}{\,dx}=0.05y with the initial condition y(0)=1000

    By separation of variables, this leads to \frac{\,dy}{y}=0.05\,dx

    Integrate both sides, and we see that \int\frac{\,dy}{y}=0.05\int\,dx\implies \ln\left|y\right|=0.05x+C\implies\left|y\right|=e^  {0.05x+C} \implies \left|y\right|=e^Ce^{0.05x}\implies y=Ke^{0.05x};~K=\pm e^{C}

    Now applying the initial condition y(0)=1000, you get 1000=Ke^{0.05\cdot0}\implies K=1000

    Thus, your function is y\!\left(x\right)=1000e^{0.05x}

    Now go ahead and find \frac{\,d^2y}{\,dx^2}

    Does this make sense? Can you take it from here?
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  3. #3
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    Second Derivative with only "y" variable- Taylor Series Needed

    Quote Originally Posted by Chris L T521 View Post
    Hmm...this is what I think you should do.

    I suggest you solve the differential equation \frac{\,dy}{\,dx}=0.05y with the initial condition y(0)=1000

    By separation of variables, this leads to \frac{\,dy}{y}=0.05\,dx

    Integrate both sides, and we see that \int\frac{\,dy}{y}=0.05\int\,dx\implies \ln\left|y\right|=0.05x+C\implies\left|y\right|=e^  {0.05x+C} \implies \left|y\right|=e^Ce^{0.05x}\implies y=Ke^{0.05x};~K=\pm e^{C}

    Now applying the initial condition y(0)=1000, you get 1000=Ke^{0.05\cdot0}\implies K=1000

    Thus, your function is y\!\left(x\right)=1000e^{0.05x}

    Now go ahead and find \frac{\,d^2y}{\,dx^2}

    Does this make sense? Can you take it from here?

    Well, yes...thank you. I wish I could do it that way, but I forgot to mention that the answer must be computed using the Taylor series method.
    I'm still not sure how to create derivatives up to the 4th power to make this equation work with Taylor series. Any suggestions?
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