Dat Do edited beginproblem__Consid.tex  almost 10 years ago

Commit id: 0def6c69580b83ad798d46346b561e2e90c890c5

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\\  Now, we will solve for $\beta$  \\  $v^{"}(n) $v^{"}_n  = \beta _1 n + \beta _2$ \\  $\beta _1 n + \beta _2 = (\beta _1(n-1)+\beta _2) + 3(\beta _1(n-2)+\beta _2) + (\beta _1(n-3) + \beta _2) + 2n$  \\ 

$5\beta _1 = -2$  \\  \\  $\beta _1 = -2/5$  $\beta _2 = -4/5$ -\frac{2}{5}$  \\  $v^{"}(n) $\beta _2  = -\frac{4}{5}$  \\  $v^{"}_n = -\frac{2}{5}n - \frac{4}{5}$  \\  General solution: $v_n = \alpha _1 (-1)^n + \alpha _2 (1+\sqrt{2})^n + \alpha _3 (1-\sqrt{2})^n  -\frac{2}{5}n - \frac{4}{5}$