what is the integral of 2t/1-t^2 ?

Question

Solve the following differential equation with the given initial condition:

(1-t^2) dy/dt = 2t, y=0 when t=0, for -1< t < 1

Answer 1

dy/dt = 2t/(1 - t²)

 

y(t) = ∫ 2t/(1 - t²) dt = -ln[1 - t²] + k where k is an integrating constant.

 

 

Solve for k using the initial condition, y(0) = 0:

 

0 = -ln[1 - 0²] + k = -ln[1] + k = 0 + k

 

So, k = 0.

 

 

Note that the domain of ln(x) is x > 0, so the domain for y would be -1 < t < 1.