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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

$${\mathrm{y}}_{\mathrm{n}+2}-{2\mathrm{y}}_{\mathrm{n}+1}+{\mathrm{y}}_{\mathrm{n}}-{\mathrm{n}}^2{2}^{\mathrm{n}}=0$$$${\mathrm{l}}^{\prime }\mathrm{equation}\mathrm{caracteristique}\mathrm{est}:$$$${\mathrm{x}}^2-2\mathrm{x}+1-{\mathrm{n}}^2{2}^{\mathrm{n}}=0$$$$\mathrm{\Delta }={\mathrm{n}}^2{2}^{\mathrm{n}+2}$$$${\mathrm{x}}_1=\frac{2-\mathrm{n}\sqrt{2^{\mathrm{n}+2}}}{2}\mathrm{et}{\mathrm{x}}_2=\frac{2+\mathrm{n}\sqrt{2^{\mathrm{n}+2}}}{2}$$$${\mathrm{d}}^{\prime }\mathrm{ou};{\mathrm{y}}_{\mathrm{n}}=\alpha {\mathrm{x}}_1^{\mathrm{n}}+\beta {\mathrm{x}}_2^{\mathrm{n}}\mathrm{with}\alpha ,\beta \in \mathbb{R}/$$$$\{\begin{array}{l}\alpha +\beta ={\mathrm{y}}_0\\ \alpha {\mathrm{x}}_1+\beta {\mathrm{x}}_2={\mathrm{y}}_1\end{array}$$

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