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?