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
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?