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Question Number 33977 by abdo imad last updated on 28/Apr/18

$$findapolynomesolutionofthediifferencial$$$$equation{y}^{{}^{″}}+y=x^{12}.$$

Answered by MJS last updated on 28/Apr/18

$$y^{″}+y=x^{2n}$$$$n=1\Rightarrow y=x^2-2=$$$$=x^2-\left(1\times 2\right)x^0$$$$n=2\Rightarrow y=x^4-12x^2+24=$$$$=x^4-\left(3\times 4\right)x^2+\left(1\times 2\times 3\times 4\right)x^0$$$$n=3\Rightarrow y=x^6-30x^4+360x^2-720=$$$$=x^6-\left(5\times 6\right)x^4+\left(3\times 4\times 5\times 6\right)x^4-\left(1\times 2\times 3\times 4\times 5\times 6\right)x^0$$$$$$$$y^{″}+y=x^{2n}\Rightarrow y=x^{2n}-\frac{\left(2n\right)!}{(2n-2)!}{x}^{2n-2}+\frac{\left(2n\right)!}{(2n-4)!}{x}^{2n-4}...\pm \left(2n\right)!x^0=$$$$=\underset{k=0}{\sum ^n}{(-1)}^k\frac{\left(2n\right)!}{(2n-2k)!}{x}^{2n-2k}$$$$$$$$y^{{}^{″}}+y=x^{12}\Rightarrow y=x^{12}-\frac{12!}{10!}{x}^{10}+\frac{12!}{8!}{x}^8-\frac{12!}{6!}{x}^6+\frac{12!}{4!}{x}^4-\frac{12!}{2!}{x}^2+12!$$

Commented by MJS last updated on 28/Apr/18

$$y^{″}+y=x^n\Rightarrow y=\underset{k=0}{\sum ^{\lfloor \frac{n}{2}\rfloor }}{(-1)}^n\frac{n!}{(n-2k)!}{x}^{n-2k}$$

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