Question

Let y be the solution of the initial value problem y''+y=sin (2 x), y(0)=0, y'(0)=0. The maximum value of y is ____.

Answer 1

The maximum value of y is 1/2.

Step-by-step explanation:

To find the solution y for the given initial value problem y'' + y = sin(2x) with y(0) = 0 and y'(0) = 0, we can start by solving the homogeneous equation y'' + y = 0. The characteristic equation for this homogeneous part is r² + 1 = 0, giving us
`r = \pm i.` The general solution for the homogeneous part is
`\(y_h(x) = A \cos(x) + B \sin(x)\),` where A and B are constants.

Now, for the particular solution, since the non-homogeneous term is (sin(2x)), we assume
`(y_p(x)\)` to be of the form
`\(y_p(x) = C \sin(2x) + D \cos(2x)\)`. Plugging this into the original differential equation, we find that C = 0and D = 1/2, leading to the particular solution
`\(y_p(x) = (1/2) \cos(2x)\).`

The general solution is the sum of the homogeneous and particular solutions
`: \(y(x) = y_h(x) + y_p(x) = A \cos(x) + B \sin(x) + (1/2) \cos(2x)\)`. Applying the initial conditions y(0) = 0 and y'(0) = 0gives A = 0 and B = 0, leaving us with the solution
`\(y(x) = (1/2) \cos(2x)\).`

The maximum value of
`\(\cos(2x)\`) is 1, so the maximum value of y is 1/2.