Answer: (i) With \(y=tx\), the differential equation becomes \(\dfrac{\mathrm d^2y}{\mathrm dt^2}+9y=3t^2+1\).
(ii) \(x=\dfrac{\cos3t-\pi\sin3t+9t^2+1}{27t}\).
Let \(y=tx\). Then
\(y'=x+tx'\)
and
\(y''=2x'+tx''.\)
The given differential equation is
\(tx''+2x'+9tx=3t^2+1.\)
Since \(tx''+2x'=y''\) and \(tx=y\), this becomes
\(y''+9y=3t^2+1.\)
Now solve
\(y''+9y=3t^2+1.\)
The complementary function is
\(y_c=A\cos3t+B\sin3t.\)
Try a particular solution \(y_p=at^2+bt+c\). Then \(y_p''=2a\), so
\(2a+9(at^2+bt+c)=3t^2+1.\)
Comparing coefficients gives
\(9a=3,\quad 9b=0,\quad 2a+9c=1.\)
Thus
\(a=\dfrac13,\quad b=0,\quad c=\dfrac1{27}.\)
So
\(y=A\cos3t+B\sin3t+\dfrac13t^2+\dfrac1{27}.\)
Since \(x=\dfrac yt\),
\(x=\dfrac{A\cos3t+B\sin3t+\frac13t^2+\frac1{27}}{t}.\)
At \(t=\dfrac\pi3\), \(x=\dfrac\pi9\), hence
\(y=tx=\dfrac\pi3\cdot\dfrac\pi9=\dfrac{\pi^2}{27}.\)
Using \(\cos\pi=-1\) and \(\sin\pi=0\),
\(-A+\dfrac{\pi^2}{27}+\dfrac1{27}=\dfrac{\pi^2}{27},\)
so
\(A=\dfrac1{27}.\)
Also
\(x'=\dfrac{ty'-y}{t^2}.\)
Now
\(y'=-3A\sin3t+3B\cos3t+\dfrac{2t}{3}.\)
At \(t=\dfrac\pi3\),
\(y'=-3B+\dfrac{2\pi}{9}.\)
Since \(x'=\dfrac23\),
\(\dfrac{\frac\pi3\left(-3B+\frac{2\pi}{9}\right)-\frac{\pi^2}{27}}{\frac{\pi^2}{9}}=\dfrac23.\)
This gives
\(-\pi B+\dfrac{\pi^2}{27}=\dfrac{2\pi^2}{27},\)
so
\(B=-\dfrac\pi{27}.\)
Therefore
\(y=\dfrac1{27}\cos3t-\dfrac\pi{27}\sin3t+\dfrac13t^2+\dfrac1{27}.\)
Finally, divide by \(t\):
\(x=\dfrac{\cos3t-\pi\sin3t+9t^2+1}{27t}.\)
