Recent content by pd1zz13

The original equation is: V'' - k*V^-1/2 = 0 I then said u = V' therefore u' = u(du/dV) so the new equation is : u(du/dV) = V^-1/2*k Im a little rusty on separation of variables but I got a u in the final answer which means I have to integrate again since I need the final answer...

ok I apologize I wrote that wrong. The original equation is: V'' - k*V^-1/2 = 0 I then said u = V' therefore u' = u(du/dV) so the new equation is : u(du/dV) = V^-1/2*k Im a little rusty on separation of variables but I got a u in the final answer which means I have to integrate...

Hi, I am trying to solve the following equation analytically. I think the solution shouldn't be that hard but I'm really rusty on these kind of things. V' - k*V^-1/2 = 0 where k is constant. Any help is appreciated, got to turn this in tomorrow!