How can we find general solution of $y(n+3)-6y(n+2)+12y(n+1)-8y(n)=0$ without using mathematica etc.
How can we find general solution of the following problem?
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ordinary-differential-equations
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0Hint: Consider the equation $$x^3 - 6x^2 +12x - 8 = 0.$$ These roots of this equation is $x_1 = x_2 = x_3 = 2$. And so the solution of the differential equation is that $$y(n) = (An^2 + Bn +C)2^n, A,B,C \in \C$$ – 2017-01-08
1 Answers
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The basic idea for difference equations like this, is to start with the ansatz $y(n) = \lambda^n$. This gives us $$ \lambda^n \cdot (\lambda^3 - 6\lambda^2 + 12 \lambda - 8) = 0 $$ Dividing by $\lambda^n$, we get $$ \lambda^3 - 6\lambda^2 + 12\lambda - 8 = (\lambda - 2)^3 = 0. $$ Hence $\lambda = 2$ is the only solution to this - so called characteristic - equation. For a triple solution of the characteristic equation we need more then the one basic solution $y_0(n) = 2^n$, we get from our ansatz. In general, if $\lambda$ is a $k$-times solution of the characteristic equations, all $n^i \lambda^n$, $0 \le i < k$ are solutions to the equation. Hence, we get the three basic solutions $$ y_i(n) = n^i 2^n, \qquad i = 0,1, 2. $$ Any solution will be a linear combination of these three, hence we get $$ y(n) = (an^2 + bn + c)2^n$$ as general solution.