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Question Number 67122 by sandeepkeshari0797@gmail.com last updated on 23/Aug/19

Answered by Cmr 237 last updated on 23/Aug/19

y_(n+2) −2y_(n+1) +y_n −n^2 2^n =0 l′equation caracteristique est: x^2 −2x+1−n^2 2^n =0 Δ=n^2 2^(n+2) x_1 =((2−n(√2^(n+2) ))/2) et x_2 =((2+n(√2^(n+2) ))/2) d′ou;y_n =αx_1 ^n +βx_2 ^n with α,β∈R/ { ((α+β=y_0 )),((αx_1 +βx_2 =y_1 )) :}

$$\mathrm{y}_{\mathrm{n}+\mathrm{2}} −\mathrm{2y}_{\mathrm{n}+\mathrm{1}} +\mathrm{y}_{\mathrm{n}} −\mathrm{n}^{\mathrm{2}} \mathrm{2}^{\mathrm{n}} =\mathrm{0} \\ $$$$\mathrm{l}'\mathrm{equation}\:\mathrm{caracteristique}\:\mathrm{est}: \\ $$$$\mathrm{x}^{\mathrm{2}} −\mathrm{2x}+\mathrm{1}−\mathrm{n}^{\mathrm{2}} \mathrm{2}^{\mathrm{n}} =\mathrm{0} \\ $$$$\Delta=\mathrm{n}^{\mathrm{2}} \mathrm{2}^{\mathrm{n}+\mathrm{2}} \\ $$$$\mathrm{x}_{\mathrm{1}} =\frac{\mathrm{2}−\mathrm{n}\sqrt{\mathrm{2}^{\mathrm{n}+\mathrm{2}} }}{\mathrm{2}}\:\mathrm{et}\:\mathrm{x}_{\mathrm{2}} =\frac{\mathrm{2}+\mathrm{n}\sqrt{\mathrm{2}^{\mathrm{n}+\mathrm{2}} }}{\mathrm{2}} \\ $$$$\mathrm{d}'\mathrm{ou};\mathrm{y}_{\mathrm{n}} =\alpha\mathrm{x}_{\mathrm{1}} ^{\mathrm{n}} +\beta\mathrm{x}_{\mathrm{2}} ^{\mathrm{n}} \:\mathrm{with}\:\alpha,\beta\in\mathbb{R}/ \\ $$$$\begin{cases}{\alpha+\beta=\mathrm{y}_{\mathrm{0}} }\\{\alpha\mathrm{x}_{\mathrm{1}} +\beta\mathrm{x}_{\mathrm{2}} =\mathrm{y}_{\mathrm{1}} }\end{cases} \\ $$

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