7. (2 points) Solve the following initial value problem (here $$ \beta $$ is a constant): $$ \begin{array}{l}\frac{d^{2} y}{d t^{2}}+y=0, \quad y(0)=\cos \beta, \quad y^{\prime}(0)=-\sin \beta\end{array} $$ (a) $$ y(t)=\cos \beta \cos t-\sin \beta \sin t $$ (b) $$ y(t)=\cos \beta \cos t+\sin \beta \sin t $$ (c) $$ y(t)=\cos \beta \sin t-\sin \beta \cos t $$ (d) $$ y(t)=\cos \beta \sin t+\sin \beta \cos t $$

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We start with the differential equation

$$
\frac{d^2 y}{dt^2} + y = 0,
$$

with the initial conditions

$$
y(0)=\cos \beta \quad \text{and} \quad y'(0)=-\sin \beta.
$$

Since this is a second-order homogeneous linear differential equation with constant coefficients, its general solution is

$$
y(t)=A\cos t+B\sin t,
$$

where $$A$$ and $$B$$ are constants determined by the initial conditions.

**Step 1: Apply the initial condition $$y(0)=\cos \beta$$.**

Substitute $$t=0$$:

$$
y(0)=A\cos 0+B\sin 0=A\cdot1+B\cdot0=A.
$$

Thus,

$$
A=\cos \beta.
$$

**Step 2: Compute the derivative of $$y(t)$$ and apply $$y'(0)=-\sin \beta$$.**

Differentiate $$y(t)$$ with respect to $$t$$:

$$
y'(t)=-A\sin t+B\cos t.
$$

Substitute $$t=0$$:

$$
y'(0)=-A\sin 0+B\cos 0=-A\cdot0+B\cdot1=B.
$$

Thus,

$$
B=-\sin \beta.
$$

**Step 3: Write the particular solution with the calculated constants.**

Substitute $$A=\cos \beta$$ and $$B=-\sin \beta$$ into the general solution:

$$
y(t)=\cos \beta\,\cos t+(-\sin \beta)\,\sin t=\cos \beta\,\cos t-\sin \beta\,\sin t.
$$

Therefore, the solution to the initial value problem is

$$
y(t)=\cos \beta\,\cos t-\sin \beta\,\sin t.
$$

This corresponds to option (a).
The solution is $$ y(t) = \cos \beta \cos t - \sin \beta \sin t $$, which matches option (a).

(2 points) Solve the following initial value problem (here is a constant):

(a)
(b)
©
(d)

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(2 points) Solve the following initial value problem (here is a constant): (a) (b) © (d)