Max Height Homogenous Differential Equation

[shards5]

Homework Statement

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

Homework Equations

N/A

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)
y_p = 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?

 

[Mark44]

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.

 

[shards5]

I made a mistake c₁ is supposed to also equal zero so my particular solution would only have 1/3sin(2x).

 

[Mark44]

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.