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Given D(n+1) =(a+b)D(n) -abD(n-1) for all n>=2 How do i find the general formula for Dn?

I already created 2 equations.

Dn = 1Dn 0
Dn+1 = (a+b)Dn -abDn-1

I found the eigenvalues 1 , -ab and corresponding eigenvectors, ((1+ab)/a+b),1), (0,1))

But now iam stuck and iam not sure what to do next

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  • $\begingroup$ \begin{eqnarray*} D_n= \alpha a^n + \beta b^n. \end{eqnarray*} $\endgroup$ Oct 26, 2019 at 14:27

1 Answer 1

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For the recurrence equation$$D_{n+1} =(a+b)D_n -abD_{n-1}$$ the characteristic equation is $$r^2=(a+b)r-ab$$ Solve the quadratic to get the roots $r_1$ and $r_2$ and use $$D_n=c_1 \,r_1^n+c_2\, r_2^n$$

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