Question

Consider the solution to the IVP 9 – = ; g (0) = 2 Find y" (0) Consider the solution to the IVP tư – (g)? = 0; }(0) = 1; / (0) = 2 Find the coefficient of 25 in its Taylor expansion centered at 0.

Answer 1

#1 The given IvP is y'-ay ; y(o)=2 y'axy=x > y = xray so y'CO) = Oto Yeo) o=co), R y'= athy 7 g" = i +(8 +ry!) .: yuco) = 1 + (460) + o y'(o)) = 1 + yo) = 1+2 =3 Again y": = 1+ y +ny! + y = y'+4+*y") = zy' + my so y" (o) = 2y.co) + 0.y"(0) = 2y (0) 11 hu. = 2X0 =0 Hence yo) #2 The given IvP is yy" -(yl)²=0; y(o)=1; y'(o)=2 The Taylor expansion of the solution yea) around azo yea) = y(0) + (2-0) plo) + (*y "CO) + (4,50% yousco) + is given by 3}" Y"CO + (9-0) (2-93 yu) (0) + 41 5) Therefore the coefficient of x5 in the Taylor expansion of y(x) is y(V) (0) 51

wir.t. x and yirco) Now yy"-(40)2=0 Ýy"=(y172 :: yogy" (o) = (440)) >> Ty" (O) = 4 yy"=(41)² * Yoghu+ yg" = 2y' (Differentiating ~ yy!"=y'y" .. 8) 8"() = 4(0) 4"(9) !yuco) = 2* 4 (putting all values of yco), y(o) > 4"(0) = yy"=yly" y'y" + y y(v) = (y")²+ y'y" (Differentiating => y yen = (y")? yeo) y) (0) = (y" (o)) you (0) = (H) = 16 → () =16 Again yyam =(yn gyly + y g(u) = 2yllyll →y Oy () +yoyco) =27'()%" () 1. ycusco) = 2x4x8 64-32 = 32 Hence the coefficient of a5 in the Taylor expansion of y(x) is 31 = w.rot. x is 2 X 16+ > Tymo 32 4 15