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Math Help - Linear Second Order DE

  1. #1
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    Linear Second Order DE

    The initial question was :

    y''-10y'+61y=0

    i've got the general solution which is e^5x(A cos(6x)+b sin(6x))

    given that the initial conditions were y(0)=0 and y'(0)=-9,find A and B

    after i calculated shudn't A =0?it seems i was wrong?

    Please help thanks!
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  2. #2
    Ted
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    A=0 is correct.

    What makes you think it is wrong?
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  3. #3
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    \displaystyle A = 0 is correct.
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  4. #4
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    then how would the general solution be like?as in y(x)=?
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  5. #5
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    Differentiate it and substitute your other condition to solve for \displaystyle B.
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  6. #6
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    I've got B= 54

    thus my equation should be 'exp(5*x)*(54*sin(6*x))'..but somehow its wrong
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  7. #7
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    I don't know how you can get \displaystyle B = 54...

    \displaystyle y'(x) = 5Be^{5x}\sin{(6x)} + 6Be^{5x}\cos{(6x)}, and since \displaystyle y'(0) = -9

    \displaystyle -9 = 5Be^{5\cdot 0}\sin{(6\cdot 0)} + 6Be^{5\cdot 0}\cos{(6\cdot 0)}

    \displaystyle -9 = 6B

    \displaystyle B = -\frac{9}{6} = -\frac{3}{2}...
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  8. #8
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    Ahh..i integrated it wrongly!!Thanks alot!
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  9. #9
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    Quote Originally Posted by topsquark View Post
    You made a slight error. A = 0, so the sine term should not be there. It worked because sin(6*0) = 0.

    -Dan
    You need to use the product rule to differentiate \displaystyle y(x) = Be^{5x}\sin{(6x)}.

    \displaystyle u(x) = Be^{5x} \implies u'(x) = 5Be^{5x} and \displaystyle v(x) = \sin{(6x)} \implies v'(x) = 6\cos{(6x)}.

    Therefore \displaystyle y'(x) = u'(x)v(x) + u(x)v'(x)

    \displaystyle = 5Be^{5x}\sin{(6x)} + 6Be^{5x}\cos{(6x)}.


    I did not make a mistake.
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  10. #10
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Prove It View Post
    I did not make a mistake.
    I....ummmm...I was just testing you. Yeah, that's it. Testing. (crawls away into a dark corner.)

    -Dan
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