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Extra Practice - Test 2: Problem 15 Previous Problem Problem List Next Problem Thl ANSWERS ONLY CHECKED -- ANSWERS NOT RECORDED Results for this submission Entered Answer Preview Result Correct Answ. Incorrect 0.866025403784 The answer above is NOT correct. (0 points) Let y be the solution of the initial value problem y"+y= - sin(2x), y(0) -0,7 (0) = 0. The maximum value of y is Show: CorrectAnswers Preview My Answers Check Answers You have attempted this problem 0 times. This homework set is closed. Email instructor

Answer 1

solution:

y'' + y = 0

λ² + 1 = 0

λ = +/- i

Characteristic solutoin:

Asin(x) + Bcos(x)

non-homogeneous solution

y'' + y = -sin2x

Potential solution to this non homogeneous solution:

y = Csin2x + Dcos2x

y' = 2Ccos2x - 2Dsin2x

y'' = -4Csin2x - 4Dcos2x

Plug this in your differential equation:

y'' + y = -sin2x

(-4Csin2x - 4Dcos2x) + Csin2x + Dcos2x = -sin2x

sin(2x) [ -4C+ C] = -1 {because we have a -sin2x term on the right side}

cos(2x) [-4D + D] = 0 { we have no cosine terms on the right side}

D = 0

C = 1/3

Solution:

y(x) = 1/3(sin(2x)) + Asin(x) + Bcos(x)

Put the initial conditions given in the solution to solve for A and B

y(0) = 0 = 0 + 0 + Bcos(0)

B = 0

y'(0) = 0 = 2/3cos(0) + A cos(0) + 0

0 = 2/3 + A

A = -2/3

Final solution:

y(x) = 1/3(sin(2x)) - 2/3sin(x)

Maximum value of this is when sin is -2pi/3 : y(x) = sqrt(3)/2

Minimum value is when sin is 2pi/3: y(x) = -sqrt(3)/2