Just wondering if anybody might be able to guide me through this problem, I'm not really sure where to start.
I wish to show that when y<<1,
τ + c ≈ [2y^(2/3)]/3 where c is a constant.
I already know the results
√(1+y) = 1+y/2 - (y^2)/8 + ....
ln[√(y)+√(1+y)] = ln[1 + √(y) + y/2 -(y^2)/8 + ....]
I originally obtained t and c from the DE
dy/dt = √(1+[1/y])
Seperation of variables and various other change of variables led me to
√(y)√(1+y) - ln[√(y)+√(1+y)] = t + c
Obviously, t and c come from the RHS (where t is a variable and c is a constant obtained from integration)
sorry, I think I'm being unclear here.
I'll start again.
Basically given dy/dt = √(1+[1/y]), I needed to show the solution was given by √(y)√(1+y) - ln[√(y)+√(1+y)] = t + c
To get there I used seperation of variables and 2 change of variables (y=q
² and then q=sinh(θ))
the next lead on question then asks me to show by series expansion that when y<<1,
t+c ≈ [2y^(2/3)]/3
I hope this has cleared things up