- #1

shards5

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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.

r

^{2}+ 1 = 0

r = 0 + i; 0- i

Thus the general solution is.

y(x) = c

_{0}cos(x) + c

_{1}sin(x)

y'(x) = -c

_{0}sin(x) + c

_{1}cos(x)

Using the initial conditions I found that.

c

_{0}= 0

c

_{1}= 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?