Q1-find-tow-power-series-solutions-of-the-given-D-E-about-x-0-y-2xy-y-0-Q2-use-the-power-series-method-to-solve-the-given-intial-value-problem-y-2xy-8y-0-y-0-3y-0-0-

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Question Number 89624 by M±th+et£s last updated on 18/Apr/20

$$Q1)findtowpowerseriessolutionsofthe$$$$givenD.Eaboutx=0$$$$y^{{}^{″}}-2{xy}^{{}^{\prime }}+y=0$$$$$$$$Q2)usethepowerseriesmethodtosolvethe$$$$givenintialvalueproblem$$$$y^{{}^{″}}-2{xy}^{{}^{\prime }}+8y=0$$$$y\left(0\right)=3y^{{}^{\prime }}\left(0\right)=0$$

Commented by mathmax by abdo last updated on 18/Apr/20

$$y^{{}^{″}}-2{xy}^{{}^{\prime }}+y=0lety=\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n\Rightarrow y^{{}^{\prime }}=\sum _{n=1}^{\mathrm{\infty }}{na}_n{x}^{n-1}=\sum _{n=0}^{\mathrm{\infty }}(n+1)a_{n+1}{x}^n$$$$and{y}^{\left(2\right)}=\sum _{n=2}^{\mathrm{\infty }}n(n-1)a_n{x}^{n-2}=\sum _{n=0}^{\mathrm{\infty }}(n+2)(n+1)a_{n+2}{x}^n$$$$\left(e\right)\Rightarrow \sum _{n=0}^{\mathrm{\infty }}(n+2)(n+1)a_{n+2}{x}^n-2\sum _{n=0}^{\mathrm{\infty }}(n+1)a_{n+1}{x}^{n+1}$$$$+\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n=0\Rightarrow$$$$\sum _{n=0}^{\mathrm{\infty }}(n+1)(n+2)a_{n+2}{x}^n-2\sum _{n=1}^{\mathrm{\infty }}{na}_n{x}^n+\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n=0$$$$\Rightarrow 2a_2+\sum _{n=1}^{\mathrm{\infty }}\{(n+1)(n+2)a_{n+2}+(1-2n)a_n\}{x}^n=0\Rightarrow$$$$\{\begin{array}{l}{a}_2=0\\ (n+1)(n+2)a_{n+2}=(2n-1)a_n\mathrm{\forall }n⩾1\Rightarrow \end{array}$$$$\{\begin{array}{l}{a}_2=0\\ a_{n+2}=\frac{2n-1}{(n+1)(n+2)}{a}_n\mathrm{\forall }n⩾1\end{array}$$$$a_{2n+2}=\frac{4n-1}{(2n+1)(2n+2)}{a}_{2n}\Rightarrow \prod _{k=2}^n\frac{a_{2k+2}}{a_{2k}}=\prod _{k=2}^n\frac{4k-1}{2(2k+1)(k+1)}$$$$\frac{a_6}{a_4}\dots \dots \frac{a_{2n}}{a_{2n-2}}.\frac{a_{2n+2}}{a_{2n}}=\frac{1}{2^n}\prod _{k=1}^n\frac{4k-1}{(k+1)(2k+1)}\Rightarrow$$$$a_{2n+2}=\frac{a_4}{2^n}(\frac{3}{2.3}\times \frac{7}{3.5}\times \frac{11}{4\times 7}\times \dots ..\frac{4n-1}{(n+1)(2n+1)})$$$$wefollowthesamewaytofind{a}_{2n+1}\dots$$

Commented by M±th+et£s last updated on 18/Apr/20

$$thanxsirandwhataboutQ2$$

Commented by mathmax by abdo last updated on 18/Apr/20

$$addtheinitalcondition\dots$$

Answered by ajfour last updated on 18/Apr/20

$$y=c_0+c_{1}x+c_2{x}^2+c_3{x}^3+\dots c_n{x}^n$$$$y^{\prime }=c_1+2c_{2}x+3c_3{x}^2+\dots .+{nc}_n{x}^{n-1}$$$$y^{″}=2c_2+2.3c_{3}x+\dots .+n(n-1)c_n{x}^{n-2}$$$$n(n-1)c_n-2(n-2)c_{n-2}+c_{n-2}=0$$$$(n+1)(n+2)c_{n+2}=(2n-1)c_n$$$$c_{n+2}=\frac{2n-1}{(n+1)(n+2)}{c}_n$$$$2c_2=c_0,c_3=\frac{c_1}{6}$$$$$$

Commented by M±th+et£s last updated on 18/Apr/20

$$thankyousir$$

Commented by M±th+et£s last updated on 18/Apr/20

$$sircanyouhelpwith{Q}_{2}andthankyou$$

Commented by ajfour last updated on 18/Apr/20

$$dontmuchrememberallthese,$$$$itsallthereinErwinKreyzig.$$

Commented by M±th+et£s last updated on 18/Apr/20

$$oksirthankyou$$

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