Given that \(y\) is a function of \(x\) and that \(x=e^u\), show that

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

Given also that

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

deduce that

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

Find \(y\) in terms of \(x\), given that \(y=0\) and \(\dfrac{dy}{dx}=-1\) when \(x=1\).