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Question Number 217797 by Tawa11 last updated on 21/Mar/25
Solve: 5x^2 y′′ + x(1 + x) y′ − y = 0
$$\mathrm{Solve}: \\ $$$$\:\:\:\:\:\mathrm{5x}^{\mathrm{2}} \:\mathrm{y}''\:\:+\:\:\:\mathrm{x}\left(\mathrm{1}\:\:+\:\:\mathrm{x}\right)\:\mathrm{y}'\:\:−\:\:\mathrm{y}\:\:\:=\:\:\:\mathrm{0} \\ $$
Answered by AntonCWX8 last updated on 22/Mar/25
I.F, μ(x)=(1/(5x^2 ))e^(∫((x(1+x))/(5x^2 ))dx) =(1/(5x^2 ))e^(∫((1+x)/(5x))dx) =(1/(5x^2 ))e^(((ln(x))/5)+(x/5)) Multiply through e^((ln(x)+x)/5) y′′ + (((1+x))/(5x))e^((ln(x)+x)/5) y′ − (1/(5x^2 ))e^((ln(x)+x)/5) y=0 Rewrite in Sturm−Liouville Form gives (e^((ln(x)+x)/5) y′)′ −(1/(5x^2 ))e^((ln(x)+x)/5) y=0 Stuck here
$${I}.{F},\:\mu\left({x}\right)=\frac{\mathrm{1}}{\mathrm{5}{x}^{\mathrm{2}} }{e}^{\int\frac{{x}\left(\mathrm{1}+{x}\right)}{\mathrm{5}{x}^{\mathrm{2}} }{dx}} =\frac{\mathrm{1}}{\mathrm{5}{x}^{\mathrm{2}} }{e}^{\int\frac{\mathrm{1}+{x}}{\mathrm{5}{x}}{dx}} =\frac{\mathrm{1}}{\mathrm{5}{x}^{\mathrm{2}} }{e}^{\frac{{ln}\left({x}\right)}{\mathrm{5}}+\frac{{x}}{\mathrm{5}}} \\ $$$${Multiply}\:{through} \\ $$$${e}^{\frac{{ln}\left({x}\right)+{x}}{\mathrm{5}}} {y}''\:+\:\frac{\left(\mathrm{1}+{x}\right)}{\mathrm{5}{x}}{e}^{\frac{{ln}\left({x}\right)+{x}}{\mathrm{5}}} {y}'\:−\:\frac{\mathrm{1}}{\mathrm{5}{x}^{\mathrm{2}} }{e}^{\frac{{ln}\left({x}\right)+{x}}{\mathrm{5}}} {y}=\mathrm{0} \\ $$$${Rewrite}\:{in}\:{Sturm}−{Liouville}\:{Form}\:{gives} \\ $$$$\left({e}^{\frac{{ln}\left({x}\right)+{x}}{\mathrm{5}}} {y}'\right)'\:−\frac{\mathrm{1}}{\mathrm{5}{x}^{\mathrm{2}} }{e}^{\frac{{ln}\left({x}\right)+{x}}{\mathrm{5}}} {y}=\mathrm{0} \\ $$$${Stuck}\:{here} \\ $$
Answered by vnm last updated on 22/Mar/25
let p be any real number y=p(x+Σ_(n=2) ^∞ (((−1)^(n−1) x^n )/(Π_(k=2) ^n (5k+1))))
$${let}\:{p}\:{be}\:{any}\:{real}\:{number} \\ $$$${y}={p}\left({x}+\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {x}^{{n}} }{\underset{{k}=\mathrm{2}} {\overset{{n}} {\prod}}\left(\mathrm{5}{k}+\mathrm{1}\right)}\right) \\ $$
Commented by Tawa11 last updated on 22/Mar/25
Please workings sir.
$$\mathrm{Please}\:\mathrm{workings}\:\mathrm{sir}. \\ $$
Answered by vnm last updated on 22/Mar/25
y=Σ_(n=0) ^∞ a_n x^n y′=Σ_(n=1) ^∞ a_n nx^(n−1) =Σ_(n=0) ^∞ a_(n+1) (n+1)x^n y′′=Σ_(n=1) ^∞ a_(n+1) (n+1)nx^(n−1) =Σ_(n=0) ^∞ a_(n+2) (n+2)(n+1)x^n xy′=Σ_(n=0) ^∞ a_(n+1) (n+1)x^(n+1) =a_1 x+Σ_(n=2) ^∞ a_n nx^n x^2 y′=Σ_(n=0) ^∞ a_(n+1) (n+1)x^(n+2) =Σ_(n=2) ^∞ a_(n−1) (n−1)x^n 5x^2 y′′=Σ_(n=0) ^∞ 5a_(n+2) (n+2)(n+1)x^(n+2) =Σ_(n=2) ^∞ 5a_n n(n−1)x^n a_0 =0 a_1 x=a_1 x ⇒ a_1 =p, p is any real number a_n n+a_(n−1) (n−1)+5a_n n(n−1)−a_n =0 a_n (n−1)+a_(n−1) (n−1)+5a_n n(n−1)=0 a_n +a_(n−1) +5a_n n=0 a_n =−(a_(n−1) /(5n+1)) a_1 =p a_2 =−(a_1 /(5∙2+1))=−(p/(11)) a_3 =−(a_2 /(5∙3+1))=(p/(11∙16)) and so on
$${y}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{n}} {x}^{{n}} \\ $$$${y}'=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{a}_{{n}} {nx}^{{n}−\mathrm{1}} =\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{n}+\mathrm{1}} \left({n}+\mathrm{1}\right){x}^{{n}} \\ $$$${y}''=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{a}_{{n}+\mathrm{1}} \left({n}+\mathrm{1}\right){nx}^{{n}−\mathrm{1}} =\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{n}+\mathrm{2}} \left({n}+\mathrm{2}\right)\left({n}+\mathrm{1}\right){x}^{{n}} \\ $$$${xy}'=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{n}+\mathrm{1}} \left({n}+\mathrm{1}\right){x}^{{n}+\mathrm{1}} ={a}_{\mathrm{1}} {x}+\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}{a}_{{n}} {nx}^{{n}} \\ $$$${x}^{\mathrm{2}} {y}'=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{n}+\mathrm{1}} \left({n}+\mathrm{1}\right){x}^{{n}+\mathrm{2}} =\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}{a}_{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right){x}^{{n}} \\ $$$$\mathrm{5}{x}^{\mathrm{2}} {y}''=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{5}{a}_{{n}+\mathrm{2}} \left({n}+\mathrm{2}\right)\left({n}+\mathrm{1}\right){x}^{{n}+\mathrm{2}} =\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\mathrm{5}{a}_{{n}} {n}\left({n}−\mathrm{1}\right){x}^{{n}} \\ $$$${a}_{\mathrm{0}} =\mathrm{0} \\ $$$${a}_{\mathrm{1}} {x}={a}_{\mathrm{1}} {x}\:\Rightarrow\:{a}_{\mathrm{1}} ={p},\:{p}\:{is}\:{any}\:{real}\:{number} \\ $$$${a}_{{n}} {n}+{a}_{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)+\mathrm{5}{a}_{{n}} {n}\left({n}−\mathrm{1}\right)−{a}_{{n}} =\mathrm{0} \\ $$$${a}_{{n}} \left({n}−\mathrm{1}\right)+{a}_{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)+\mathrm{5}{a}_{{n}} {n}\left({n}−\mathrm{1}\right)=\mathrm{0} \\ $$$${a}_{{n}} +{a}_{{n}−\mathrm{1}} +\mathrm{5}{a}_{{n}} {n}=\mathrm{0} \\ $$$${a}_{{n}} =−\frac{{a}_{{n}−\mathrm{1}} }{\mathrm{5}{n}+\mathrm{1}} \\ $$$${a}_{\mathrm{1}} ={p} \\ $$$${a}_{\mathrm{2}} =−\frac{{a}_{\mathrm{1}} }{\mathrm{5}\centerdot\mathrm{2}+\mathrm{1}}=−\frac{{p}}{\mathrm{11}} \\ $$$${a}_{\mathrm{3}} =−\frac{{a}_{\mathrm{2}} }{\mathrm{5}\centerdot\mathrm{3}+\mathrm{1}}=\frac{{p}}{\mathrm{11}\centerdot\mathrm{16}} \\ $$$${and}\:{so}\:{on} \\ $$
Commented by Tawa11 last updated on 23/Mar/25
Thanks sir.
$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$

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