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I have a question statement like this:

Show that solution of

$ y'' + NK\frac{y'}{t_f-t} + NK^2\frac{y}{(t_f-t)^2}=0 $

is

$ y(t) = C_1(t-t_f) + C_2(t-t_f)^N $.

N, K and tf are constants.
C1 and C2 are arbitrary constants.
y(t) is defined in the interval [0,tf).

I can't find a way to solve this differential equation. Can you please guide me by showing me a starting point. Any idea will be appreciated.

(Note: I double checked that I correctly wrote the question statement.)

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    hi hkBattousai, actually, it should be zero with ANY choice of C1 and C2, not just "a correct choice." But it should still be straightforward to check (the annoying part is keeping track which letters are constants).2012-12-19

1 Answers 1

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This should be a comment, but it is too long.

I would like to start noting that the answer is not correct. Taking the solution $(t_f-t)$ and substituting on the equation we get $(NK^2-NK)/(t_f-t)$, which is not zero.

In case you want also a method of solution for some equations of this type (for more general ones, exponentiation methods give always the solution, and I am sure there are more astucious methods than this one), consider a fuction that has the form: $ y(t)g(t) $ if we derive it twice and make it be zero, we get: $ y''(t)g(t)+2y'(t)g'(t)+y(t)g''(t)=0 $ or equivalently (outside zeros and etc): $ y''(t)+\frac{2g'(t)}{g(t)}y'(t)+y(t)\frac{g''(t)}{g(t)}=0. $

Comparing it with and equation of the form: $ y''(t)+2Ay'(t)+By(t)=0. $ we have: $ \frac{g'(t)}{g(t)} = \frac{A}{t_f-t} \text{ and } \frac{g''(t)}{g(t)} = \frac{B}{(t_f-t)^2} $

Deriving the first equation, and comparing with the second, to have a solution using this method we must have: $ A^2+A=B $ in which case we solve the first equation as usual to obtain: $ g(t) = C(t-t_f)^{-A} $

The equation $(yg)''=0$ can be solved easily as a linear function, and from that we get the solution: $ y(t) = C_1(t-t_f)^A + C_2(t-t_f)^{A+1} $

But I remember, this does not solve all equations of this type. A good way to see is that it only gives two solutions that differ in degree by one!