Suppose \(f\) is continuous on \((a, \infty),\) where \(a<0,\) so \(x_{0}=0\) is in
\((a, \infty)\).
(a) Use variation of parameters to find a formula for the solution of the
initial value problem
$$
y^{\prime \prime}+y=f(x), \quad y(0)=k_{0}, \quad y^{\prime}(0)=k_{1}.
$$
HINT: You will need the addition formulas for the sine and cosine:
$$
\begin{array}{l}
\sin (A+B)=\sin A \cos B+\cos A \sin B \\
\cos (A+B)=\cos A \cos B-\sin A \sin B.
\end{array}
$$
For the rest of this exercise assume that the improper integral
\(\int_{0}^{\infty} f(t) d t\) is absolutely convergent.
(b) Show that if \(y\) is a solution of
$$
y^{\prime \prime}+y=f(x)
$$
on \((a, \infty),\) then
$$
\lim _{x \rightarrow \infty}\left(y(x)-A_{0} \cos x-A_{1} \sin x\right)=0
$$
and
$$
\lim _{x \rightarrow \infty}\left(y^{\prime}(x)+A_{0} \sin x-A_{1} \cos
x\right)=0
$$
where
$$
A_{0}=k_{0}-\int_{0}^{\infty} f(t) \sin t d t \quad \text { and } \quad
A_{1}=k_{1}+\int_{0}^{\infty} f(t) \cos t d t .
$$
(c) Show that if \(A_{0}\) and \(A_{1}\) are arbitrary constants, then there's a
unique solution of \(y^{\prime \prime}+y=\) \(f(x)\) on \((a, \infty)\) that
satisfies \((\mathrm{B})\) and \((\mathrm{C})\)
