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Math Help - Laplace Transformations

  1. #1
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    Laplace Transformations

    Solve using Laplace Transformations

    1. y' + y = sin(x) where  y(0) = 1

    sY(s) - y(0) + 4Y(s) = 1/(s^2 + 1)
    Y(s)(s + 4) = 1/(s^2 + 1) - 1
    Y(s) = (s^2 + 2)/[(s + 4)(s^2 +1)]



    2. y^(^3^) + y' = e^x  where y(0)=y'(0)=y"(0)=0

    s^3Y(s) - s^2y(0) - sy'(0) - y"(0) + Y(s) - y(0) = 1/(s - 1)
    Y(s)(s^3 + s) = 1/(s - 1)
    Y(s) = 1/[x(x + 1)(x^2 + 1)]

    I'd greatly appreciate it if someone could assist me in finishing these problems as I'm unsure of what to do.
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  2. #2
    Behold, the power of SARDINES!
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    You have a slight error in the first one

    y'+y=\sin(x), \quad y(0)=1

    Taking the Laplace transform gives

    sY-1+Y=\frac{1}{s^2+1} \iff (s+1)Y=1+\frac{1}{s^2+1}

    Y=\frac{1}{s+1}+\frac{1}{(s+1)(s^2+1)}=\frac{s^2+2  }{(s+1)(s^2+1)}

    Now from here you need to use partial fractions on

    \frac{s^2+2}{(s+1)(s^2+1)}=\frac{\frac{3}{2}}{s+1}  +\frac{1}{2}\cdot\frac{-s+1}{s^2+1}=\frac{3}{2}\cdot \frac{1}{s+1}-\frac{1}{2}\cdot \frac{s}{s^2+1}+\frac{1}{2}\cdot \frac{1}{s^2+1}

    Each of these should be known transforms so the inverse is

    \frac{3}{2}e^{-t}-\frac{1}{2}\cos(t)+\frac{1}{2}\sin(t)

    for the 2nd one the first two lines are correct but the third should be

    Y=\frac{1}{s(s^2+1)(s-1)}

    Now use partial fractions to find the inverse transform.
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  3. #3
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    Y(s) = 1/(s(s^2 + 1)(s - 1))
    Y(s) = (1/2)*1/(s-1) - 1/s + (1/2)*s/(s^2 + 1) - 1/2*(s^2 + 1)
    y = (1/2)e^x - 1 + (1/2)cos(x) - (1/2)sin(x)

    Is that the right answer for the second problem?
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  4. #4
    Behold, the power of SARDINES!
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    Quote Originally Posted by Naples View Post
    Y(s) = 1/(s(s^2 + 1)(s - 1))
    Y(s) = (1/2)*1/(s-1) - 1/s + (1/2)*s/(s^2 + 1) - 1/2*(s^2 + 1)
    y = (1/2)e^x - 1 + (1/2)cos(x) - (1/2)sin(x)

    Is that the right answer for the second problem?
    Yes, Also remember you can take some derivative and plug back into the ODE to check that the answer is correct.
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