midtlinjeg

2021-02-12

Use Taylor series expansion to ﬁnd the third degree power series solution of the initial value problem
$\left(x-1\right)y\prime \prime +2y\prime -4y=0,y\left(0\right)=2,y\prime \left(0\right)=6$

Tasneem Almond

Given that:
$\left(x-1\right)y\prime \prime +2y\prime -4y=0,y\left(0\right)=2,y\prime \left(0\right)=6$
Let $y=\sum _{n=0}^{\mathrm{\infty }}{a}_{n}{x}^{n}$ be a power series expansion of the required solution.
$y=\sum _{n=0}^{\mathrm{\infty }}{a}_{n}{x}^{n}$
${y}^{\prime }=\sum _{n=0}^{\mathrm{\infty }}{a}_{n}\left(n\right){x}^{n-1}={a}_{1}+2{a}_{2}x+3{a}_{3}{x}^{2}+...$
${y}^{″}=\sum _{n=0}^{\mathrm{\infty }}{a}_{n}n\left(n-1\right){x}^{n-2}=2{a}_{2}+6{a}_{3}x+12{a}_{4}{x}^{2}+20{a}_{5}{x}^{3}+...$
Now, find the Taylor series expansions of the given series:
$\left(x-1\right)y\prime \prime +2y\prime -4y=0,y\left(0\right)=2,y\prime \left(0\right)=6$
$y\left(0\right)=2⇒{a}_{0}=2$
${y}^{\prime }\left(0\right)=6⇒{a}_{1}=6$
$\left(x-1\right)y\prime \prime +2y\prime -4y=0$

$⇒\left(x-1\right)\left(2{a}_{2}+6{a}_{3}x+12{a}_{4}{x}^{2}+20{a}_{5}{x}^{3}+...\right)+2\left({a}_{1}+2{a}_{2}x+3{a}_{3}{x}^{2}+...\right)-4\left({a}_{0}+{a}_{1}x+{a}_{2}{x}^{2}+{a}_{3}{x}^{3}+...\right)=0$$⇒\left(2{a}_{2}x+6{a}_{3}{x}^{2}+12{a}_{4}{x}^{3}+...\right)-\left(2{a}_{2}+6{a}_{3}x+12{a}_{4}{x}^{2}+20{a}_{5}{x}^{3}+...\right)+\left(2{a}_{1}+4{a}_{2}x+6{a}_{3}{x}^{2}+...\right)-\left(4{a}_{0}+4{a}_{1}x+4{a}_{2}{x}^{2}+4{a}_{3}{x}^{3}+...\right)=0$
$⇒\left(2{a}_{2}x+6{a}_{3}{x}^{2}+12{a}_{4}{x}^{3}+...\right)-\left(2{a}_{2}+6{a}_{3}x+12{a}_{4}{x}^{2}+20{a}_{5}{x}^{3}+...\right)+\left(12+4{a}_{2}x+6{a}_{3}{x}^{2}+...\right)-\left(8+24x+4{a}_{2}{x}^{2}+4{a}_{3}{x}^{3}+...\right)=0$
Compare the coefficient of ${x}^{0},{x}^{1}$ and power:
$\left(-2{a}_{2}+12-8\right)+\left(2{a}_{2}-6{a}_{3}+4{a}_{2}-24\right)x+...$
$-2{a}_{2}+12-8=0⇒-2{a}_{2}=-12+8⇒{a}_{2}=2$
Coefficient of x:
$2{a}_{2}-6{a}_{3}+4{a}_{2}-24=0$
$2{a}_{2}-6{a}_{3}+4{a}_{2}=24$

Jeffrey Jordon