Answer: (a) \(y=2e^{\frac{x}{2}}-2e^{-\frac{x}{2}}-3\).
(b) When \(y=0\), \(x=\ln 4\).
(a) Solve the homogeneous equation \(4\dfrac{d^2y}{dx^2}-y=0\).
Using a trial solution \(y=e^{mx}\), the auxiliary equation is
\(4m^2-1=0\),
so \(m=\pm \dfrac12\).
Hence the complementary function is
\(y=Ae^{-\frac{x}{2}}+Be^{\frac{x}{2}}\).
For a particular solution, try a constant \(y=k\). Then \(y''=0\), so substituting into \(4y''-y=3\) gives
\(-k=3\),
hence \(k=-3\).
So the general solution is
\(y=Ae^{-\frac{x}{2}}+Be^{\frac{x}{2}}-3\).
Differentiate:
\(\dfrac{dy}{dx}=-\dfrac12Ae^{-\frac{x}{2}}+\dfrac12Be^{\frac{x}{2}}\).
Use the conditions at \(x=0\):
From \(y(0)=-3\):
\(A+B-3=-3\Rightarrow A+B=0\).
From \(y'(0)=2\):
\(-\dfrac12A+\dfrac12B=2\).
Solving gives \(A=-2\) and \(B=2\).
Therefore
\(y=2e^{\frac{x}{2}}-2e^{-\frac{x}{2}}-3\).
(b) Set \(y=0\):
\(2e^{\frac{x}{2}}-2e^{-\frac{x}{2}}-3=0\).
So
\(2\left(e^{\frac{x}{2}}-e^{-\frac{x}{2}}\right)=3\).
Using \(\sinh u=\dfrac{e^u-e^{-u}}{2}\) with \(u=\dfrac{x}{2}\),
\(4\sinh\left(\dfrac{x}{2}\right)=3\).
Hence
\(\dfrac{x}{2}=\sinh^{-1}\left(\dfrac34\right)\).
Now use
\(\sinh^{-1}(t)=\ln\left(t+\sqrt{t^2+1}\right)\).
Then
\(\dfrac{x}{2}=\ln\left(\dfrac34+\sqrt{\dfrac{9}{16}+1}\right)=\ln\left(\dfrac34+\dfrac54\right)=\ln 2\).
Therefore
\(x=2\ln 2=\ln 4\).
