Answer: The required solution is \(y=2\ln x+1-\dfrac{1}{x}\bigl(\cos(4\ln x)+\sin(4\ln x)\bigr)\).

Let \(x=e^u\), so \(u=\ln x\). Since \(\frac{du}{dx}=\frac{1}{x}\), the chain rule gives

\(\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=\frac{1}{x}\frac{dy}{du}\).

Hence

\(x\frac{dy}{dx}=\frac{dy}{du}.\)

For the second derivative,

\(\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{1}{x}\frac{dy}{du}\right).\)

Using the product rule and \(\frac{du}{dx}=\frac{1}{x}\),

\(\frac{d^2y}{dx^2}=-\frac{1}{x^2}\frac{dy}{du}+\frac{1}{x}\frac{d}{dx}\left(\frac{dy}{du}\right) =-\frac{1}{x^2}\frac{dy}{du}+\frac{1}{x}\frac{d^2y}{du^2}\cdot\frac{du}{dx}\).

So

\(\frac{d^2y}{dx^2}=\frac{1}{x^2}\frac{d^2y}{du^2}-\frac{1}{x^2}\frac{dy}{du}\),

and therefore

\(x^2\frac{d^2y}{dx^2}=\frac{d^2y}{du^2}-\frac{dy}{du}.\)

Now substitute these into

\(x^2\frac{d^2y}{dx^2}+3x\frac{dy}{dx}+17y=34\ln x+21.\)

Since \(\ln x=u\), this becomes

\(\left(\frac{d^2y}{du^2}-\frac{dy}{du}\right)+3\frac{dy}{du}+17y=34u+21,\)

so

\(\frac{d^2y}{du^2}+2\frac{dy}{du}+17y=34u+21.\)

We now solve this ODE in \(u\). The complementary equation is

\(m^2+2m+17=0,\)

so

\(m=-1\pm 4i.\)

Thus

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

For a particular solution, try \(y_p=cu+d\). Then \(y_p'=c\) and \(y_p''=0\), so

\(2c+17(cu+d)=34u+21.\)

Matching coefficients gives \(17c=34\), hence \(c=2\), and then \(2c+17d=21\), so \(4+17d=21\) and \(d=1\). Therefore

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

Since \(u=\ln x\) and \(e^{-u}=\frac{1}{x}\),

\(y=\frac{1}{x}\big(A\cos(4\ln x)+B\sin(4\ln x)\big)+2\ln x+1.\)

Use the condition \(y=0\) when \(x=1\). Then \(\ln 1=0\), so

\(0=A+1,\)

hence \(A=-1\).

Differentiate with respect to \(x\):

\(\frac{dy}{dx}=-\frac{1}{x^2}\big(A\cos(4\ln x)+B\sin(4\ln x)\big) +\frac{1}{x}\left(-A\sin(4\ln x)\cdot\frac{4}{x}+B\cos(4\ln x)\cdot\frac{4}{x}\right)+\frac{2}{x}.\)

When \(x=1\), this gives

\(-1=-A+4B+2.\)

With \(A=-1\),

\(-1=1+4B+2,\)

so \(4B=-4\) and \(B=-1\).

Therefore

\(y=2\ln x+1-\frac{1}{x}\big(\cos(4\ln x)+\sin(4\ln x)\big)\).