Question

(x^2)y″+(5x)y′+4y=x^6, y(1)=8, y′(1)=0. y = ?

Answer 1

Gireen initial realue probleme is d n²y" & suylt 4y = 16 . 4011=8 1'111=0 net & n=e2 z=logy de dia Now, dy y & de aut 1 de la using these in me i fy". yy) 454(44) + 4y = 262 a y" _glf sy + 4y = por

as y't uy t4y = eoz consides only the homogeneous part, - ausiliary efecatron ics - m2fumth=0 3 cm +21220 an--२ । 2 Therefore complementary Function is CF = (a + Cg Z) e 22 barticular solution P-I = (e6z. 02440+ - = esz 6²448644 - est 36f2uty - l est प Therefore ylz) = oF & P. I 0 = (a + GZ) e 22 + 1 esz 64 = (at slegse) ep logo + esregie - 64 = (4+q logne) e lage) y(x) = faite logre) free 64 22 using initial conditions. you = 8 2 8 = a + 1/4 a 4 = 8-164 =

- - __y'sh - +9-4 celega ___y.c17 2050 = 3 -2ch + _ 42 २८, -1- 3 ____५८ 127 fyini = ( +ps ere) + * २५ -१ 32 ।। 32 3 32- 0 08 32 + /27 648
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