Question

Consider the initial value problem below has a series solution centered at zero of y = $\phi$ (x). Determine $\phi$ '(0), $\phi$ ''(0) and $\phi$ ⁴(0).

y''+ x²y'+ cos(x)y = 0, y(0) = 2, y'(0) = 3.

Answer 1

Soin: Given, initial value problem which has a series Solution centered) al mero of y=0(). y's xy'+ cos(x)y=0; y(0)= a, y'(o)-3 => Cosa: Ž com os > Let yw= 4a) Žaga hºo (an)! na => you. an nla"-) & yea) - Žandco-) 322 By replacing the values of y(x), y"C), yx) &, cosa, we get o Žannen as) 2Žannai, ŽE (3) h: 2 => 242 + 6azxt 1204824 2025 23+ 3000 7 't ...... + a,x?+ 22, 234 3az a't...... + 20 - af 2,227 , 4: 26-ig az at.... Therefore, (200 tao) + 6032+ Cia9449,)+ Capag 7222-2.07 (Boac + Baz)29+.... 0 = 292+ ao=0 y az=0 => 129449,=0 > 2008 + 29 2 -1 -2 , = 0

by Applying values we get Ylo)= 2 40=2 & y/o) = 3 9,=3 Therefore, we get 20-2; , = 3 ; Qg= -1; 93=0; 94--4 ; 95= 70-... Hence, $cx): Quta, at agx?4 a3 234 Q x 94 95 m2 tomon. pica) = 21+ d2224 3az a' } 404234 595 *** Ølx) = 243+ 6 az 2 + 120721 20agat... = Gaz +24042+ 600 - 2404+ 2005?f... So, values $60): 913 3 ff" Co): dag= -2 $360): 623= 0 $9400) = 24045-6