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Max Height Homogenous Differential Equation

  1. Dec 3, 2009 #1
    1. The problem statement, all variables and given/known data
    Let y be the the solution of the initial value problem. The maximum value of y is?
    y" + y = -sin(2x)
    y(0) = 0
    y'(0) = 0

    2. Relevant equations

    N/A

    3. The attempt at a solution
    Factor out original equation to find general equation.
    r2 + 1 = 0
    r = 0 + i; 0- i
    Thus the general solution is.
    y(x) = c0cos(x) + c1sin(x)
    y'(x) = -c0sin(x) + c1cos(x)
    Using the initial conditions I found that.
    c0 = 0
    c1 = 1
    Now to find A*cos(2x) + Bsin(2x)
    yp = A*cos(2x) + B*sin(2x)
    y'p = -2A*sin(2x) + 2B*cos(2x)
    y"p = -4A*cos(2x) - 4B*sin(2x)
    Plugging into the original equation we get.
    -4A*cos(2x) - 4B*sin(2x) + A*cos(2x) + B*sin(2x) = -sin(2x)
    which gives us with -3Bsin(2x) = -sin(2x) B = 1/3 and A = 0
    Therefore the equation is.
    y(x) = 1/3*sin(2x)
    To find the maximum value I do the derivative of the above equation and set it equal to zero.
    y'(x) = 2/3*cos(2x) = 0 which gives x to be pi/4
    Using x = pi/4 in the unique equation.
    y(pi/4) = 1/3*sin(2*pi/4)
    I get y = 0.333333333333333 which is wrong, of course.
    So what am I doing wrong?
     
  2. jcsd
  3. Dec 3, 2009 #2

    Mark44

    Staff: Mentor

    You're working only with the particular solution, not the general solution. Your general solution is y(x) = sin(x) + (1/3)sin(2x). Find the maximum value of that function.
     
  4. Dec 3, 2009 #3
    I made a mistake c1 is supposed to also equal zero so my particular solution would only have 1/3sin(2x).
     
  5. Dec 3, 2009 #4

    Mark44

    Staff: Mentor

    The initial conditions are for the general solution y(x), not just for the complementary solution. Start with a general solution of y(x) = c0cosx + c1sinx + (1/3)sin2x and calculate the constants c0 and c1 using your two initial conditions.
     
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