I have the following question from my textbook:
Suppose that a simple pendulum has the length $l$ of it's string increased at a steady rate (for example, a weight swaying as it is lowered by a crane). Find the equation of motion for small oscillations.
For a simple pendulum; it's equation of motion is given by $$\dfrac{d}{dt}\left(ml^2\dot\theta\right)+mgl\sin\theta=0\tag{1}$$ Let the length $l$ of the string at time $t$ be $$l=l_0+vt\tag{2}$$ and change from $t$ to $l$ as the independent variable. For small oscillations, we may replace $\sin \theta$ by $\theta$.
Hence show that equation $(1)$ becomes $$l\dfrac{d^2\theta}{dl^2}+2\dfrac{d\theta}{dl}+\dfrac{g}{v^2}\theta=0\tag{3}$$ Hint: From equation $(2)$ $dl=v\,dt$, so $$\dfrac{d}{dt}=v\dfrac{d}{dl}$$
My attempt:
Using the hint, equation $(1)$ becomes $$v\dfrac{d}{dl}\left(ml^2\dot\theta\right)+mgl\theta=0$$ $$\implies v\dfrac{d}{dl}\left(l^2\frac{d\theta}{dt}\right)+gl\theta=0$$ $$\implies v\left(2l\frac{d\theta}{dl}+l^2\frac{d^2\theta}{dl^2}\cdot v\right)+gl\theta=0\tag{a}$$ Where in $(\mathrm{a})$ I used the product rule along with the chain rule for the second term, namely $$\frac{d}{dl}\frac{d\theta}{dt}=\frac{d}{dl}\frac{d\theta}{dl}\frac{dl}{dt}=\frac{d^2\theta}{dl^2}\cdot v$$ Simplifying $(\mathrm{a})$ gives $$2v\frac{d\theta}{dl}+lv^2\frac{d^2\theta}{dl^2}+g\theta=0$$ $$\implies l\frac{d^2\theta}{dl^2}+2\frac{1}{v}\frac{d\theta}{dl}+\frac{g}{v^2}\theta=0$$ which is identical to $(3)$ with the exception of the $\dfrac{1}{v}$ factor in the second term.
I have already checked the errata list for the book and there is no error. So this leads me to believe I am making a mistake somewhere.
Is there any chance someone could point out the error in my calculation?
