Answer: (a) Since \(x=\mathrm{e}^{u}\), we have \(u=\ln x\) and therefore \(\dfrac{du}{dx}=\dfrac{1}{x}\).
So
\(\dfrac{dy}{dx}=\dfrac{dy}{du}\dfrac{du}{dx}=\dfrac{1}{x}\dfrac{dy}{du}\),
hence \(x\dfrac{dy}{dx}=\dfrac{dy}{du}\).
Differentiating again with respect to \(x\),
\(\dfrac{d}{dx}\left(x\dfrac{dy}{dx}\right)=\dfrac{d}{dx}\left(\dfrac{dy}{du}\right)\).
The left-hand side is \(x\dfrac{d^2y}{dx^2}+\dfrac{dy}{dx}\), while the right-hand side is
\(\dfrac{d}{du}\left(\dfrac{dy}{du}\right)\dfrac{du}{dx}=\dfrac{1}{x}\dfrac{d^2y}{du^2}\).
Therefore
\(x\dfrac{d^2y}{dx^2}+\dfrac{dy}{dx}=\dfrac{1}{x}\dfrac{d^2y}{du^2}\),
and multiplying by \(x\) gives
\(x^2\dfrac{d^2y}{dx^2}+x\dfrac{dy}{dx}=\dfrac{d^2y}{du^2}\).
Using \(x\dfrac{dy}{dx}=\dfrac{dy}{du}\), this becomes
\(x^2\dfrac{d^2y}{dx^2}=\dfrac{d^2y}{du^2}-\dfrac{dy}{du}\).
(b) Substitute these relations into \(x^2\dfrac{d^2y}{dx^2}+5x\dfrac{dy}{dx}+3y=30x^2\):
\(\left(\dfrac{d^2y}{du^2}-\dfrac{dy}{du}\right)+5\dfrac{dy}{du}+3y=30\mathrm{e}^{2u}\).
Hence
\(\dfrac{d^2y}{du^2}+4\dfrac{dy}{du}+3y=30\mathrm{e}^{2u}\).
To solve, first find the complementary function from
\(m^2+4m+3=0\), so \((m+1)(m+3)=0\) and \(m=-1,-3\).
Thus
\(y_c=A\mathrm{e}^{-u}+B\mathrm{e}^{-3u}\).
For a particular integral, try \(y_p=Ce^{2u}\). Then
\(y_p'=2Ce^{2u}\), \(y_p''=4Ce^{2u}\).
Substituting gives
\((4C+8C+3C)e^{2u}=30e^{2u}\), so \(15C=30\) and \(C=2\).
Therefore the general solution in terms of \(u\) is
\(y=A\mathrm{e}^{-u}+B\mathrm{e}^{-3u}+2\mathrm{e}^{2u}\).
Finally, since \(\mathrm{e}^{u}=x\), we have \(\mathrm{e}^{-u}=x^{-1}\), \(\mathrm{e}^{-3u}=x^{-3}\), and \(\mathrm{e}^{2u}=x^2\). So
\(y=\dfrac{A}{x}+\dfrac{B}{x^3}+2x^2\).
Let \(x=\mathrm{e}^{u}\), so \(u=\ln x\) and \(\dfrac{du}{dx}=\dfrac{1}{x}\).
Then
\(\dfrac{dy}{dx}=\dfrac{dy}{du}\dfrac{du}{dx}=\dfrac{1}{x}\dfrac{dy}{du}\),
so \(x\dfrac{dy}{dx}=\dfrac{dy}{du}\).
Differentiating \(x\dfrac{dy}{dx}=\dfrac{dy}{du}\) with respect to \(x\),
\(\dfrac{d}{dx}\left(x\dfrac{dy}{dx}\right)=\dfrac{d}{dx}\left(\dfrac{dy}{du}\right)\).
The left-hand side is \(x\dfrac{d^2y}{dx^2}+\dfrac{dy}{dx}\).
For the right-hand side, use the chain rule:
\(\dfrac{d}{dx}\left(\dfrac{dy}{du}\right)=\dfrac{d}{du}\left(\dfrac{dy}{du}\right)\dfrac{du}{dx}=\dfrac{1}{x}\dfrac{d^2y}{du^2}\).
So
\(x\dfrac{d^2y}{dx^2}+\dfrac{dy}{dx}=\dfrac{1}{x}\dfrac{d^2y}{du^2}\).
Multiplying by \(x\) gives
\(x^2\dfrac{d^2y}{dx^2}+x\dfrac{dy}{dx}=\dfrac{d^2y}{du^2}\).
Since \(x\dfrac{dy}{dx}=\dfrac{dy}{du}\), we obtain
\(x^2\dfrac{d^2y}{dx^2}=\dfrac{d^2y}{du^2}-\dfrac{dy}{du}\).
Now substitute into \(x^2\dfrac{d^2y}{dx^2}+5x\dfrac{dy}{dx}+3y=30x^2\):
\(\left(\dfrac{d^2y}{du^2}-\dfrac{dy}{du}\right)+5\dfrac{dy}{du}+3y=30\mathrm{e}^{2u}\),
hence
\(\dfrac{d^2y}{du^2}+4\dfrac{dy}{du}+3y=30\mathrm{e}^{2u}\).
To solve this differential equation, first find the complementary function from
\(m^2+4m+3=0\).
So \((m+1)(m+3)=0\), giving \(m=-1,-3\).
Therefore
\(y_c=A\mathrm{e}^{-u}+B\mathrm{e}^{-3u}\).
For a particular solution, try \(y_p=Ce^{2u}\). Then
\(y_p'=2Ce^{2u}\), \(y_p''=4Ce^{2u}\).
Substituting into the equation gives
\((4C+8C+3C)e^{2u}=30e^{2u}\), so \(15C=30\) and \(C=2\).
Thus
\(y=A\mathrm{e}^{-u}+B\mathrm{e}^{-3u}+2\mathrm{e}^{2u}\).
Finally, since \(\mathrm{e}^{u}=x\),
\(\mathrm{e}^{-u}=\dfrac{1}{x}\), \(\mathrm{e}^{-3u}=\dfrac{1}{x^3}\), and \(\mathrm{e}^{2u}=x^2\).
So the general solution in terms of \(x\) is
\(y=\dfrac{A}{x}+\dfrac{B}{x^3}+2x^2\).
