Integral of $(d^2y/dx^2) dy$

I saw this as a step in a calculation and it was confusing. The left cancels down to:

$$\int \frac{d^2y}{dx^2}dy$$

But surely the answer is $\frac{d^2y}{dx^2}y$; instead, $\frac {d^2}{dx^2}$ is being treated as a constant and not $\frac {d^2y}{dx^2} $

. Putting in the calculation directly yields the result I would have expected.

1 Answer

For the left side you have that: $$I=\int y''y'dx= \frac 12\int(y'^2)'dx=\frac 12 \int \dfrac {d(y')^2}{dx}dx$$ $$I=\frac 12 \int d(y')^2=\frac 12 (y')^2$$

Note that; $$(y'^2)'=2y'y''$$