Max Height Homogenous Differential Equation

  • Thread starter shards5
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  • #1
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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)
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?
 

Answers and Replies

  • #2
35,139
6,892
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.
 
  • #3
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I made a mistake c1 is supposed to also equal zero so my particular solution would only have 1/3sin(2x).
 
  • #4
35,139
6,892
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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