Help with Fluid Dynamics question

Question

Got this question as part of a worksheet from my fluid dynamics module. I know that I need to solve $dx/dt=y$ and $dy/dt= -(x-bt)$ but I don't know how to get it into the form with $\sin$ and $\cos$. Any help at all?

The offending question

Answer 1

Given that

$\dfrac{dx}{dt} = y \tag 1$

and

$\dfrac{dy}{dt} = -(x - bt) = - x + bt, \tag 2$

we may differentiate (2) to obtain

$\dfrac{d^2y}{dt^2} = -\dfrac{dx}{dt} + b; \tag 3$

if we substitute (1) into (3):

$\dfrac{d^2y}{dt^2} = -y + b, \tag 4$

or

$\dfrac{d^2y}{dt^2} + y = b; \tag 5$

we are given

$x(0) = x_0, \tag 6$

and

$y(0) = y_0; \tag 7$

therefore, from (2),

$\dfrac{dy(0)}{dt} = -(x(0) - b \cdot 0) = -x_0; \tag 8$

we now note that the solution to (5) with initial conditions (7), (8) is

$y(t) = (y_0 - b)\cos t - x_0 \sin t + b; \tag 9$

again from (2),

$x(t)= -\dfrac{dy(t)}{dt} + bt = (y_0 - b) \sin t + x_0 \cos t + bt, \tag{10}$

in agreement with the given solutions.