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Thread: 2nd order DE

  1. #1
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    2nd order DE

    I have an equation y''+y=0 with y(0)=1 and y'(0)=0

    It has general solution of y=Asin(x)+Bcos(x) ?

    with condition gives 1 = 0 + B so B=1

    If so y' = Acos(x)-sin(x)

    with condition gives 0 = A - 0 so A=0

    Therefore the solution is y = cos(x)

    Am I going about this the write way?


    And if so does anything change to the general solution if I am solving

    y''+4y=0 with y(0)=1 and y'(0)=0 ?

    Thank you!
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  2. #2
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    Quote Originally Posted by Bushy View Post
    I have an equation y''+y=0 with y(0)=1 and y'(0)=0

    It has general solution of y=Asin(x)+Bcos(x) ?

    with condition gives 1 = 0 + B so B=1

    If so y' = Acos(x)-sin(x)

    with condition gives 0 = A - 0 so A=0

    Therefore the solution is y = cos(x)

    Am I going about this the write way?
    Yes!


    And if so does anything change to the general solution if I am solving

    y''+4y=0 with y(0)=1 and y'(0)=0 ?
    Yes - $\displaystyle
    y = A \sin 2x +B \cos 2x
    $
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  3. #3
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    Quote Originally Posted by Danny View Post
    Yes - $\displaystyle
    y = A \sin 2x +B \cos 2x
    $
    Is this because in general differentiating sin(2x) and cos(2x) twice will give me -4sin(2x) and -4cos(2x) ?
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  4. #4
    Senior Member Twig's Avatar
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    Hi!

    If you have $\displaystyle y''+4y=0 $ , you have what is known as the characteristic equation.

    In this case it becomes $\displaystyle r^{2}+4=0 $ , which clearly has no real solutions. You get $\displaystyle r^{2}=-2 $ , which gives $\displaystyle r=\pm 2i $

    So the general solution becomes $\displaystyle A\cdot cos(2x) + B\cdot sin(2x)$ .

    In general, if you get a solution to the characteristic equation, in the form

    $\displaystyle a+bi $ , then the homogenous solution is $\displaystyle e^{ax}(Acos(bx)+Bsin(bx))$
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  5. #5
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    Thank you, this makes sense.
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