Question

Solve the given initial-value problem.
d2x/dt2 + 4x = −2 sin(2t) + 8 cos(2t),

x(0) = −1, x'(0) = 1

Answer 1

Combining the homogeneous and the particular solutions, and applying the initial conditions, the solution to the given IVP is:

$x(t) = -\cos{(2t)} + \frac{1}{2}\sin{(2t)}$

The IVP is given by:

$x^{\prime\prime} + 4x = -2\sin{(2t)} + 8\cos{(2t)}$

First, we find the homogeneous solution, which depends on the roots of the characteristic equation, given by:

$r^2 + 4 = 0$

$r^2 = -4$

$r = \pm \sqrt{-4}$

$r = \pm 2i$

Thus:

$x_h(t) = a\cos{(2t)} + b\sin{(2t)}$

Then, according to the source(left-side of the equality), the particular solution has the following format:

$x_p(t) = A\cos{(2t)} + B\sin{(2t)}$

Applying the derivatives:

$x_p^{\prime}(t) = -2A\sin{(2t)} + 2B\cos{(2t)}$

$x_p^{\prime\prime}(t) = -4A\cos{(2t)} - 4B\sin{(2t)}$

Replacing into the IVP:

$x^{\prime\prime} + 4x = -2\sin{(2t)} + 8\cos{(2t)}$

$-4A\cos{(2t)} - 4B\sin{(2t)} + 4A\cos{(2t)} + 4B\sin{(2t)} = -2\sin{(2t)} + 8\cos{(2t)}$

$0A\cos{(2t)} + 0B\sin{(2t)} = -2\sin{(2t)} + 8\cos{(2t)}$

Thus, the particular solution is $x_p(t) = 0$.

The complete solution is:

$x(t) = x_h(t) + x_p(t)$

$x(t) = x_h(t)$

$x(t) = A\cos{(2t)} + B\sin{(2t)}$

Applying the initial conditions:

$x(0) = -1 \rightarrow A + 0B = -1 \rightarrow A = -1$

$x^{\prime}(0) = 1 \rightarrow 2B = 1 \rightarrow B = \frac{1}{2}$

Thus, the solution is:

$x(t) = -\cos{(2t)} + \frac{1}{2}\sin{(2t)}$

A similar problem is given at https://brainly.com/question/13143214