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I have a recurrence relation for a fund that starts a 50 million, 6 % interest every year, and an outtake of 2 million/year. How to find out a solution for what funds exists after n years?

$$x_{n+1}=x_{n}*1.06-2$$ $$ x_{0}=50$$

Under are my calculations -

Assosiated homogenous relation : $$x_{n+1}=1.06x_{n}$$

Special solution : $$x_{n+1}=1.06x_{n}-2 = x_{n}^{s}=33.3334$$

General solution : $$x_{n}=C*(1.06)^n+33.3334$$

How to calculate the last unknown C? The answer in the book says its 50/3, but i really can't get my head around why its not just 50 as the $x_{0}$ is.

(Sorry in advance if i have misspelled some of the solution names or the equation part names, i've tried translating them from norwegian. Help on correct translations is appreciated.)

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    $\begingroup$ First write 33.3334= 100/3. Then x0 = C + 100/3 it is to say 50 = C + 100/3. Then C = 50/3. Do you agree? $\endgroup$ – Juan Ospina Sep 29 '14 at 19:00
  • $\begingroup$ Your line with the "Special solution" is not clear. You'd better write "a constant value $x_{n+1}=x_n=K$, such that $K=1.06K-2$, works". $\endgroup$ – Yves Daoust Sep 29 '14 at 19:11

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