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Question Number 143766 by akmalovna05 last updated on 18/Jun/21
x×y′′−y=x^� 3
$${x}×{y}''−{y}=\hat {{x}}\mathrm{3} \\ $$
Commented by TheHoneyCat last updated on 18/Jun/21
what is x^∧ 3 ? do you mean: x×y′′−y=x^3 ?
$${what}\:{is}\:{x}^{\wedge} \mathrm{3}\:? \\ $$$$ \\ $$$${do}\:{you}\:{mean}:\:{x}×{y}''−{y}={x}^{\mathrm{3}} \:? \\ $$
Answered by Olaf_Thorendsen last updated on 18/Jun/21
xy′′−y = x^3 (1) y = Σ_(n=0) ^∞ a_n x^n y′′ = Σ_(n=2) ^∞ n(n−1)a_n x^(n−2) = Σ_(n=0) ^∞ (n+2)(n+1)x^n (1) : xΣ_(n=0) ^∞ (n+2)(n+1)a_(n+2) x^n −Σ_(n=0) ^∞ a_n x^n = x^3 Σ_(n=0) ^∞ (n+2)(n+1)a_(n+2) x^(n+1) −Σ_(n=0) ^∞ a_n x^n = x^3 Σ_(n=1) ^∞ n(n+1)a_(n+1) x^n −Σ_(n=0) ^∞ a_n x^n = x^3 { ((−a_0 = 0)),((1.2.a_2 −a_1 = 0)),((2.3.a_3 −a_2 = 0)),((3.4.a_4 −a_3 = 1)),((n(n+1)a_(n+1) −a_n = 0 (n>3))) :} { ((a_0 = 0)),((a_2 = (a_1 /2))),((a_3 = (a_2 /6) = (a_1 /(12)))),((a_4 = ((a_3 +1)/(12)) = (a_1 /(144))+(1/(12)))),((a_(n+1) = (a_n /(n(n+1))) (n>3))) :} a_(n+1) = (a_n /(n(n+1))) (n>3) a_(n+1) = (a_(n−1) /((n+1)n^2 (n−1))) a_(n+1) = (a_(n−2) /((n+1)n^2 (n−1)^2 (n−2))) a_(n+1) = (a_(n−2) /((n+1)n^2 (n−1)^2 (n−2)^2 ...5^2 .4)) a_(n+1) = (a_4 /((n+1)n^2 (n−1)^2 (n−2)^2 ...5^2 .4)) a_(n+1) = ((3!4!a_4 )/(n!(n+1)!)) a_(n+1) = ((24a_4 )/(n!(n+1)!)) a_(n+1) = ((24((a_1 /(144))+(1/(12))))/(n!(n+1)!)) a_(n+1) = (((a_1 /6)+2)/(n!(n+1)!)) y = a_1 x+(a_1 /2)x^2 +(a_1 /(12))x^3 +((a_1 /6)+2)Σ_(n=4) ^∞ (x^n /(n!(n+1)!)) a_1 is any constant
$${xy}''−{y}\:=\:{x}^{\mathrm{3}} \:\:\:\:\:\left(\mathrm{1}\right) \\ $$$${y}\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{n}} {x}^{{n}} \\ $$$${y}''\:=\:\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}{n}\left({n}−\mathrm{1}\right){a}_{{n}} {x}^{{n}−\mathrm{2}} \:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left({n}+\mathrm{2}\right)\left({n}+\mathrm{1}\right){x}^{{n}} \\ $$$$\left(\mathrm{1}\right)\::\:{x}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left({n}+\mathrm{2}\right)\left({n}+\mathrm{1}\right){a}_{{n}+\mathrm{2}} {x}^{{n}} −\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{n}} {x}^{{n}} \:=\:{x}^{\mathrm{3}} \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left({n}+\mathrm{2}\right)\left({n}+\mathrm{1}\right){a}_{{n}+\mathrm{2}} {x}^{{n}+\mathrm{1}} −\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{n}} {x}^{{n}} \:=\:{x}^{\mathrm{3}} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{n}\left({n}+\mathrm{1}\right){a}_{{n}+\mathrm{1}} {x}^{{n}} −\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{n}} {x}^{{n}} \:=\:{x}^{\mathrm{3}} \\ $$$$ \\ $$$$\begin{cases}{−{a}_{\mathrm{0}} \:=\:\mathrm{0}}\\{\mathrm{1}.\mathrm{2}.{a}_{\mathrm{2}} −{a}_{\mathrm{1}} \:=\:\mathrm{0}}\\{\mathrm{2}.\mathrm{3}.{a}_{\mathrm{3}} −{a}_{\mathrm{2}} \:=\:\mathrm{0}}\\{\mathrm{3}.\mathrm{4}.{a}_{\mathrm{4}} −{a}_{\mathrm{3}} \:=\:\mathrm{1}}\\{{n}\left({n}+\mathrm{1}\right){a}_{{n}+\mathrm{1}} −{a}_{{n}} \:=\:\mathrm{0}\:\left({n}>\mathrm{3}\right)}\end{cases} \\ $$$$\begin{cases}{{a}_{\mathrm{0}} \:=\:\mathrm{0}}\\{{a}_{\mathrm{2}} \:=\:\frac{{a}_{\mathrm{1}} }{\mathrm{2}}}\\{{a}_{\mathrm{3}} \:=\:\frac{{a}_{\mathrm{2}} }{\mathrm{6}}\:=\:\frac{{a}_{\mathrm{1}} }{\mathrm{12}}}\\{{a}_{\mathrm{4}} \:=\:\frac{{a}_{\mathrm{3}} +\mathrm{1}}{\mathrm{12}}\:=\:\frac{{a}_{\mathrm{1}} }{\mathrm{144}}+\frac{\mathrm{1}}{\mathrm{12}}}\\{{a}_{{n}+\mathrm{1}} \:=\:\frac{{a}_{{n}} }{{n}\left({n}+\mathrm{1}\right)}\:\left({n}>\mathrm{3}\right)}\end{cases} \\ $$$${a}_{{n}+\mathrm{1}} \:=\:\frac{{a}_{{n}} }{{n}\left({n}+\mathrm{1}\right)}\:\left({n}>\mathrm{3}\right) \\ $$$${a}_{{n}+\mathrm{1}} \:=\:\frac{{a}_{{n}−\mathrm{1}} }{\left({n}+\mathrm{1}\right){n}^{\mathrm{2}} \left({n}−\mathrm{1}\right)} \\ $$$${a}_{{n}+\mathrm{1}} \:=\:\frac{{a}_{{n}−\mathrm{2}} }{\left({n}+\mathrm{1}\right){n}^{\mathrm{2}} \left({n}−\mathrm{1}\right)^{\mathrm{2}} \left({n}−\mathrm{2}\right)} \\ $$$${a}_{{n}+\mathrm{1}} \:=\:\frac{{a}_{{n}−\mathrm{2}} }{\left({n}+\mathrm{1}\right){n}^{\mathrm{2}} \left({n}−\mathrm{1}\right)^{\mathrm{2}} \left({n}−\mathrm{2}\right)^{\mathrm{2}} …\mathrm{5}^{\mathrm{2}} .\mathrm{4}} \\ $$$${a}_{{n}+\mathrm{1}} \:=\:\frac{{a}_{\mathrm{4}} }{\left({n}+\mathrm{1}\right){n}^{\mathrm{2}} \left({n}−\mathrm{1}\right)^{\mathrm{2}} \left({n}−\mathrm{2}\right)^{\mathrm{2}} …\mathrm{5}^{\mathrm{2}} .\mathrm{4}} \\ $$$${a}_{{n}+\mathrm{1}} \:=\:\frac{\mathrm{3}!\mathrm{4}!{a}_{\mathrm{4}} }{{n}!\left({n}+\mathrm{1}\right)!} \\ $$$${a}_{{n}+\mathrm{1}} \:=\:\frac{\mathrm{24}{a}_{\mathrm{4}} }{{n}!\left({n}+\mathrm{1}\right)!} \\ $$$${a}_{{n}+\mathrm{1}} \:=\:\frac{\mathrm{24}\left(\frac{{a}_{\mathrm{1}} }{\mathrm{144}}+\frac{\mathrm{1}}{\mathrm{12}}\right)}{{n}!\left({n}+\mathrm{1}\right)!} \\ $$$${a}_{{n}+\mathrm{1}} \:=\:\frac{\frac{{a}_{\mathrm{1}} }{\mathrm{6}}+\mathrm{2}}{{n}!\left({n}+\mathrm{1}\right)!} \\ $$$$ \\ $$$${y}\:=\:{a}_{\mathrm{1}} {x}+\frac{{a}_{\mathrm{1}} }{\mathrm{2}}{x}^{\mathrm{2}} +\frac{{a}_{\mathrm{1}} }{\mathrm{12}}{x}^{\mathrm{3}} +\left(\frac{{a}_{\mathrm{1}} }{\mathrm{6}}+\mathrm{2}\right)\underset{{n}=\mathrm{4}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} }{{n}!\left({n}+\mathrm{1}\right)!} \\ $$$${a}_{\mathrm{1}} \:\mathrm{is}\:\mathrm{any}\:\mathrm{constant} \\ $$

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