Integrating $ \frac{\mathrm{d}^{2}v}{\mathrm{d}y^{2}} = \frac{\mathrm{d}p}{\mathrm{d}x} $

Question

How would I go about integrating this to figure out what $\mathrm{d}v/\mathrm{d}y$ is? The bounds on $y$ is ($H$--> constant upper, and $y$ varied lower). I know how to generally do it but I'm not sure how to formally write it out.

Answer 1

We integrate both sides with respect to $y$:

$\int_y^H\frac{d^2v}{d^2y}dy=\int_y^H\frac{dp}{dx}dy$

$\left.\frac{dv}{dy}\right|_{y=H}-\frac{dv}{dy}=(H-y)\frac{dp}{dx}$

We integrate both sides with respect to $y$ again:

$\int_y^H\left(\left.\frac{dv}{dy}\right|_{y=H}-\frac{dv}{dy}\right)dy=\int_y^H(H-y)\frac{dp}{dx}dy$

$(H-y)\left.\frac{dv}{dy}\right|_{y=H}-\left.v\right|_{y=H}+v=\frac{dp}{dx}\left(\frac{y^2+H^2}{2}-yH\right)$