Answer: The general solution is \(y=e^{-x/2}(A\sin 4x+B\cos 4x)+x^2+1\), where \(A\) and \(B\) are arbitrary constants.

Hence \(\dfrac{y}{x^2}=\dfrac{e^{-x/2}(A\sin 4x+B\cos 4x)}{x^2}+1+\dfrac{1}{x^2}\), so \(\dfrac{y}{x^2}\to 1\) as \(x\to\infty\), whatever the initial conditions.

We first solve the homogeneous equation

\(4y''+4y'+65y=0\).

Try a solution of the form \(y=e^{rx}\). This gives the auxiliary equation

\(4r^2+4r+65=0\).

So

\(r=\dfrac{-4\pm\sqrt{16-1040}}{8}=\dfrac{-4\pm\sqrt{-1024}}{8}=\dfrac{-4\pm 32i}{8}=-\dfrac12\pm 4i\).

Therefore the complementary function is

\(y_c=e^{-x/2}(A\sin 4x+B\cos 4x)\).

Now find a particular integral. Since the right-hand side is a quadratic, try

\(y_p=ax^2+bx+c\).

Then \(y_p'=2ax+b\) and \(y_p''=2a\).

Substitute into the differential equation:

\(4(2a)+4(2ax+b)+65(ax^2+bx+c)=65x^2+8x+73\).

Collecting coefficients gives

\(65a\,x^2+(8a+65b)x+(8a+4b+65c)=65x^2+8x+73\).

Equating coefficients:

\(65a=65\Rightarrow a=1\),

\(8a+65b=8\Rightarrow 8+65b=8\Rightarrow b=0\),

\(8a+4b+65c=73\Rightarrow 8+65c=73\Rightarrow c=1\).

So a particular integral is \(y_p=x^2+1\).

Hence the general solution is

\(y=e^{-x/2}(A\sin 4x+B\cos 4x)+x^2+1\).

To show the limiting result, divide by \(x^2\):

\(\dfrac{y}{x^2}=e^{-x/2}\left(\dfrac{A\sin 4x+B\cos 4x}{x^2}\right)+1+\dfrac{1}{x^2}.\)

As \(x\to\infty\), the exponential term tends to \(0\), and the factor \(\dfrac{A\sin 4x+B\cos 4x}{x^2}\) is bounded in magnitude by \(\dfrac{|A|+|B|}{x^2}\), which also tends to \(0\). Also \(\dfrac{1}{x^2}\to 0\).

Therefore \(\dfrac{y}{x^2}\to 1\) as \(x\to\infty\), whatever the initial conditions.