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Thread: 2ODE Linear with one constant?

  1. #1
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    2ODE Linear with one constant?

    $\displaystyle y^{\prime \prime} + 25 y = 0$

    Initial Conditions are:
    $\displaystyle y(0)=-1,\quad y(\frac{\pi}{10})=-2$

    So, My roots are $\displaystyle 0\pm10i$


    Its in the form P+- qi
    Does the fact that my P=0 change anything? Or should I just roll with it?

    Anyways, heres what I did

    $\displaystyle Y=ACos(5x)+Bsin(5x))$
    $\displaystyle Y'=-A5sin(5x)+B5Cos(5x)$

    So, when Y(0)=-1
    We get $\displaystyle A=-1$

    When $\displaystyle Y'(\frac{\pi}{10})=-2$
    we get:
    $\displaystyle -2=-A5$ which is just a false statement.
    Im not able to isolate "B" given the initial conditions. What am I doing wrong?
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  2. #2
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    I'm pretty sure that the roots aren't $\displaystyle \displaystyle 0 \pm 10i$...
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  3. #3
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    Ur right, Its 0 +- 5i

    Thats the value I actually used, I just copied it here wrong.
    But still, my problem persists!
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  4. #4
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Vamz View Post
    $\displaystyle y^{\prime \prime} + 25 y = 0$

    Initial Conditions are:
    $\displaystyle y(0)=-1,\quad y(\frac{\pi}{10})=-2$

    So, My roots are $\displaystyle 0\pm10i$


    Its in the form P+- qi
    Does the fact that my P=0 change anything? Or should I just roll with it?

    Anyways, heres what I did

    $\displaystyle Y=ACos(5x)+Bsin(5x))$
    $\displaystyle Y'=-A5sin(5x)+B5Cos(5x)$

    So, when Y(0)=-1
    We get $\displaystyle A=-1$

    When $\displaystyle Y'(\frac{\pi}{10})=-2$
    we get:
    $\displaystyle -2=-A5$ which is just a false statement.
    Im not able to isolate "B" given the initial conditions. What am I doing wrong?
    You differentiated y(x) for the condition y(pi/10) which you didnt need to.

    ie just use $\displaystyle Y=ACos(5x)+Bsin(5x))$ for both conditions
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