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Math Help - 2nd order diff Eq

  1. #1
    cheeee
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    2nd order diff Eq

    How to solve this second order diff eq, need help!?
    2500y''+9y=0;
    y(0) = 1
    y'(0) = 1
    I think you have to substitute with z = dy/dt and reduce it to a separable equation..but somehow the math is still not working out for me.
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  2. #2
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    Hello, cheeee!

    This is not a "separable" problem . . .


    Solve: . 2500y''+9y\:=\:0,\quad<br />
y(0) \,= \,1,\;\;y'(0) \,=\, 1
    The characteristic equation is: . 2500m^2 + 9 \:=\:0
    . . which has roots: .  m \:=\:\pm\frac{3}{50}i

    Hence, the general solution is: . y \;=\;C_1\cos\left(\frac{3}{50}x\right) + C_2\sin\left(\frac{3}{50}x\right)


    Since y(0) = 1\!:\;\;C_1\cos(0) + C_2\sin(0) \:=\:1\quad\Rightarrow\quad\boxed{ C_1 \:=\:1}


    Differentiate: . y' \;=\;-\frac{3}{50}C_1\sin\left(\frac{3}{50}x\right) + \frac{3}{50}C_2\cos\left(\frac{3}{50}x\right)

    Since y'(0)=1\!:\;\;-\frac{3}{50}C_1\sin(0) + \frac{3}{50}C_2\cos(0) \:=\:1\quad\Rightarrow\quad\boxed{ C_2 = \frac{50}{3}}


    Therefore: . \boxed{\;y \;=\;\cos\left(\frac{3}{50}x\right) + \frac{50}{3}\sin\left(\frac{3}{50}x\right)\;}

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  3. #3
    Behold, the power of SARDINES!
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    Quote Originally Posted by cheeee View Post
    How to solve this second order diff eq, need help!?
    2500y''+9y=0;
    y(0) = 1
    y'(0) = 1
    I think you have to substitute with z = dy/dt and reduce it to a separable equation..but somehow the math is still not working out for me.
    50^2y''+3^2y=0

    solve the equation

    50^2m^2+3^2=0

    rewriting we get

    (50m)^2+3^2=(50m)^2-(3i)^2=(50m-3i)(50m+3i)=0


    so the two solutions for m are

    m=\frac{3i}{50} or =\frac{-3i}{50}

    so we end up with the complimentry solution.

    y_c=c_1\cos(\frac{3}{50}t)+c_2\sin{\frac{3}{50}t}
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