Help with some fiddly algebra around a 1st order ODE

[ognik]

Frustratingly although I can solve the ODE, I am getting a different answer to the book. Now going in circles so would appreciate a fresh pair of eyes.

The ODE (for a boat coasting with resistance proportional to $V^n$) starts as $ m\frac{dV}{dt} =-kV^n $ Find V(t) and x(t), V(0) = $V_0$

I solved the general case for n as $ V={V}_{0}[1+(n-1)\frac{k}{m}t{V}_{0}^{n-1}]^{\frac{1}{1-n}} $, which happily agrees with the book. But I simplified the power of the [...] term to $ V={V}_{0}[1+(n-1)\frac{k}{m}t{V}_{0}^{n-1}]^{n-1} $ - can't see that would make a difference?

Then, integrating again to get x(t), I let the term [...] = u, then $ dt=\frac{m}{(n-1)k{V}_{0}^{n-1}}du $

and $ x(t)=\frac{m{V}_{0}}{(n-1)k{V}_{0}^{n-1}}\int{u}^{n-1}du = \frac{m{V}_{0}}{(n-1)k{V}_{0}^{n-1}}\frac{{u}^{n}}{n} = \frac{m{V}_{0}^{2-n}}{kn(n-1)}[...]^n $

But the book shows $ \frac{m{V}_{0}^{2-n}}{k(2-n)}(1 - [...]^\frac{n-2}{n-1}) $?

 

[Ackbach]

Frustratingly although I can solve the ODE, I am getting a different answer to the book. Now going in circles so would appreciate a fresh pair of eyes.

The ODE (for a boat coasting with resistance proportional to $V^n$) starts as $ m\frac{dV}{dt} =-kV^n $ Find V(t) and x(t), V(0) = $V_0$

I solved the general case for n as $ V={V}_{0}[1+(n-1)\frac{k}{m}t{V}_{0}^{n-1}]^{\frac{1}{1-n}} $, which happily agrees with the book. But I simplified the power of the [...] term to $ V={V}_{0}[1+(n-1)\frac{k}{m}t{V}_{0}^{n-1}]^{n-1} $ - can't see that would make a difference?

Why is
$$\frac{1}{1-n}=n-1?$$
That's the only thing you've changed in the expression. But this equality is manifestly not true. If for no other reason, $n=1$ is not in the domain of the LHS, but it is in the RHS.

 

[ognik]

How I can get the hard stuff done and then do something like that is beyond my understanding... maybe it's time to take a break, I think I mixed up $a^{-n} $ as $ a^{\frac{1}{n}}$ instead of $\frac{1}{{a}^{n}} $ - there is no simplification of a power like $ \frac{1}{1-n}$ is there?

 

[Ackbach]

Hey, it happens to the best of us. Algebra mistakes are by far the most common I run across, even in advanced math courses.

No, I don't think you can simplify the exponent at all. I think you're stuck with it.

Cheers!