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Math Help - Second Order Nonhomogeneous Equation

  1. #1
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    Second Order Nonhomogeneous Equation

    Hey everyone, I need help understanding solving this equation as it deals with an input function that is discontinuous, which I don't know how to handle.

    g(x) =sin x for 0 \le x \le \frac{\pi}{2}
    g(x)=0 for x > \frac{\pi}{2}
    Solve y''+4y=g(x)

    The two intervals:
    y'' +4y = sin (x) for 0 \le x \le \frac{\pi}{2}

    and

    y'' +4y = 0 for x > \frac{\pi}{2}

    with y(0)= 1 and y'(0)=2

    Can anyone help me?

    Thanks
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  2. #2
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    Solve the homogeneous equation .

    It has characteristic equation , which has solution .

    From this, we can determine that the solution to the homogeneous DE is . This is also the entire solution when .


    Now for the nonhomogeneous solution, i.e. when , we try . This would mean and .

    Substituting into the DE gives

    .



    So the particular solution is .

    Therefore when , the solution of the DE is .

    Now you will need to use your boundary conditions to evaluate A and B.
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  3. #3
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Prove It View Post
    Solve the homogeneous equation .

    It has characteristic equation , which has solution .

    From this, we can determine that the solution to the homogeneous DE is . This is also the entire solution when .
    May I ask how do you know this is the entire solution when you dont know A and B at this stage?
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  4. #4
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    Because g(x) = 0 when x > pi/2, which means the DE is homogeneous in that domain.
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  5. #5
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    Quote Originally Posted by Nguyen View Post
    Hey everyone, I need help understanding solving this equation as it deals with an input function that is discontinuous, which I don't know how to handle.

    g(x) =sin x for 0 \le x \le \frac{\pi}{2}
    g(x)=0 for x > \frac{\pi}{2}
    Solve y''+4y=g(x)

    The two intervals:
    y'' +4y = sin (x) for 0 \le x \le \frac{\pi}{2}

    and

    y'' +4y = 0 for x > \frac{\pi}{2}

    with y(0)= 1 and y'(0)=2

    Can anyone help me?

    Thanks
    Since this is an IVP an alternative is to use the Laplace transform to solve note that

    \sin(x)=\cos\left( x-\frac{\pi}{2}\right)

    So the right hand side can be written as

    g(x)=\sin(x)-u\left(x-\frac{\pi}{2} \right)\cos\left( x-\frac{\pi}{2}\right)

    and this has Laplace transform

    G(s)=\frac{1}{s^2+1}-e^{-\frac{\pi}{2}s}\cdot \frac{s}{s^2+1}
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  6. #6
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    Thanks Prove It and TheEmptySet, I did the question the same way as Prove It but crossed it out becuase I didn't know how to deal with g(x) being discontinuous.

    I got y= 5/6 sin(2x) +cos(2x)+1/3 sin(x)
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  7. #7
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    Quote Originally Posted by Nguyen View Post
    Thanks Prove It and TheEmptySet, I did the question the same way as Prove It but crossed it out becuase I didn't know how to deal with g(x) being discontinuous.

    I got y= 5/6 sin(2x) +cos(2x)+1/3 sin(x)
    Prove It's method was simply telling you to do the problem over two domains. Not that complicated! I'd recommend learning to do it that way. It's good practice.

    -Dan
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