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Thread: Solve this 2nd order IVP!!

  1. #1
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    Exclamation Solve this 2nd order IVP!!

    I missed the day of class where we went over solving 2nd order diff-eqs.

    y''(t) - 100 y(t) = 0

    subject to y(0)=100 and y'(0)=3000

    Thank you!
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  2. #2
    Junior Member slovakiamaths's Avatar
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    Quote Originally Posted by tanelly View Post
    y''(t) - 100 y(t) = 0
    subject to y(0)=100 and y'(0)=3000
    integrate both sides
    $\displaystyle y'-50y^2=a$................(i)
    again integrate
    $\displaystyle y-\frac{50}{3}y^3=ay+b$.......................(ii)
    now put t=0 in eq (i)&(ii)
    3000-50(10000)=a
    100-(50/3)*1000000=100a+b
    solve it for a and b and pun in eq (ii) to get the req ans
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  3. #3
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    Quote Originally Posted by tanelly View Post
    I missed the day of class where we went over solving 2nd order diff-eqs.

    y''(t) - 100 y(t) = 0

    subject to y(0)=100 and y'(0)=3000

    Thank you!
    $\displaystyle y''(t) - 100y(t) = 0$

    This is second order linear constant coefficient homogeneous.
    So the characteristic equation is

    $\displaystyle m^2 - 100 = 0$

    $\displaystyle m^2 = 100$

    $\displaystyle m = \pm 10$.


    So the solution is

    $\displaystyle y(t) = c_1e^{10t} + c_2e^{-10t}$.


    You have two initial conditions: $\displaystyle y(0) = 100$ and $\displaystyle y'(0) = 3000$.

    From the first, we can see

    $\displaystyle 100 = c_1e^{10(0)} + c_2e^{-10(0)}$

    $\displaystyle 100 = c_1 + c_2$.


    From the second...

    $\displaystyle y(t) = c_1e^{10t} + c_2e^{-10t}$

    $\displaystyle y'(t) = 10c_1e^{10t} - 10c_2e^{-10t}$

    $\displaystyle 3000 = 10c_1e^{10(0)} - 10c_2e^{-10(0)}$

    $\displaystyle 3000 = 10c_1 - 10c_2$

    $\displaystyle 300 = c_1 - c_2$.


    So now you have two equations in two unknowns:

    $\displaystyle 100 = c_1 + c_2$

    $\displaystyle 300 = c_1 - c_2$.


    Adding them together, we find

    $\displaystyle 100 + 300 = (c_1 + c_2) + (c_1 - c_2)$

    $\displaystyle 400 = 2c_1$

    $\displaystyle c_1 = 200$.


    Back substituting:

    $\displaystyle 100 = c_1 + c_2$

    $\displaystyle 100 = 200 + c_2$

    $\displaystyle c_2 = -100$.



    So finally, the solution is

    $\displaystyle y(t) = 200e^{10t} - 100e^{-10t}$.
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  4. #4
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    Quote Originally Posted by slovakiamaths View Post
    integrate both sides
    $\displaystyle y'-50y^2=a$................(i)
    again integrate
    $\displaystyle y-\frac{50}{3}y^3=ay+b$.......................(ii)
    now put t=0 in eq (i)&(ii)
    3000-50(10000)=a
    100-(50/3)*1000000=100a+b
    solve it for a and b and pun in eq (ii) to get the req ans
    Sorry, but this won't work.

    $\displaystyle y''(t) - 100y(t) = 0$

    $\displaystyle y''(t) = 100y(t)$.


    If you were going to integrate both sides, you would have to integrate with respect to $\displaystyle t$, not $\displaystyle y$. And since you don't know anything about $\displaystyle y$ yet, you can't possibly integrate it with respect to $\displaystyle t$.
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