Answer: \(y=\left[e^{-x}\left(\frac12\sin2x-\cos2x\right)+3+2x-x^2\right]^{-1}\).

Let \(v=\frac1y\). Then

\(\frac{dv}{dx}=-\frac1{y^2}\frac{dy}{dx}\),

and

\(\frac{d^2v}{dx^2}=-\frac1{y^2}\frac{d^2y}{dx^2}+\frac2{y^3}\left(\frac{dy}{dx}\right)^2\).

Substituting these expressions into the given equation gives

\(\frac{d^2v}{dx^2}+2\frac{dv}{dx}+5v=17+6x-5x^2\), as required.

Now solve the linear differential equation. The complementary equation has auxiliary equation

\(m^2+2m+5=0\),

so \(m=-1\pm2i\). Hence

\(v_c=e^{-x}(A\cos2x+B\sin2x)\).

For a particular integral, try \(v_p=px^2+qx+r\). Then \(v_p'=2px+q\) and \(v_p''=2p\). Substitution gives

\(2p+2(2px+q)+5(px^2+qx+r)=17+6x-5x^2\).

Equating coefficients gives \(5p=-5\), \(4p+5q=6\), and \(2p+2q+5r=17\). Therefore \(p=-1\), \(q=2\), and \(r=3\).

So

\(v=e^{-x}(A\cos2x+B\sin2x)+3+2x-x^2\).

When \(x=0\), \(y=\frac12\), so \(v=2\). Hence \(2=A+3\), giving \(A=-1\).

Also, since \(v=\frac1y\),

\(\frac{dv}{dx}=-\frac1{y^2}\frac{dy}{dx}\).

At \(x=0\), \(y=\frac12\) and \(\frac{dy}{dx}=-1\), so \(\frac{dv}{dx}=4\).

Differentiate \(v\):

\(\frac{dv}{dx}=e^{-x}(-2A\sin2x+2B\cos2x)-e^{-x}(A\cos2x+B\sin2x)+2-2x\).

At \(x=0\), this gives \(4=2B-A+2\). With \(A=-1\), \(4=2B+3\), so \(B=\frac12\).

Thus

\(v=e^{-x}\left(\frac12\sin2x-\cos2x\right)+3+2x-x^2\).

Since \(y=\frac1v\),

\(y=\left[e^{-x}\left(\frac12\sin2x-\cos2x\right)+3+2x-x^2\right]^{-1}\).