Separation of Variables: Solving M(dv/dt) = [k*(v^2)] - Mg

[Richard Platt]

Homework Statement

Help solving this separation of variables. M(dv/dt) = [k*(v^2)] - Mg

Homework Equations

As above

The Attempt at a Solution

dv/(kv^2-Mg)= dt/M

Intregrate both sides;

(1/2kv) ln |kv^2-Mg|= t/m + c?

Then what to do..? Something to do with with Ae^(-kt/M)..?

 

[Mark44]

Homework Statement

Help solving this separation of variables. M(dv/dt) = [k*(v^2)] - Mg

Homework Equations

As above

The Attempt at a Solution

dv/(kv^2-Mg)= dt/M

Intregrate both sides;

(1/2kv) ln |kv^2-Mg|= t/m + c?

Your integration is wrong here. You apparently thought you were working with $\int du/u$, but you weren't. If u = kv² - Mg, then du = 2kvdv, but you can't just stick in a factor of v as you seem to have done.

To integrate dv/(kv² - Mg), break the denominator up using partial fractions. Alternatively, you could use a trig substitution, but partial fractions would probably be easier.

Then what to do..? Something to do with with Ae^(-kt/M)..?

P.S. Welcome to PF!

 

[dextercioby]

In the LHS a substitution $v=\sqrt{\frac{Mg}{k}} \cosh p$ would help you solve the integral, assuming $k,m,g > 0$.