Question

ind the solution of the given initial value problem:

6y′′′+144y′−960y=0

y(0)=9, y′(0)=42, y′′(0)=−240

Enclose arguments of functions in parentheses. For example, sin(2x).

Answer 1

given initiala value problem LINP) is byll +1449) aboy = 0; 410/=9, ylo)=42, y" (0) = -2.40 Now, by +14.4.191- qboy ao - yll+244-Iboy = 0 Ausillary Equalion to is m² +24m-160 =D clearly to mah is root 24 -160 1bo n4 1Ь 21-0 24 matima 40 =0. -> m² +2.2.m +2²+36=0 > (m+2²=-36 - mx2=tbi > ma-2abi Hence roots to e are m=4,-2 tbt Genesal solution is ly=9e42e2" (ce costsalt cg sim(ba)). y'all ciety te-PalcecoslbxJ+Cg sinuj] =>y'= 49,642_ 2E 2x [Cacoolbalty sinltas) + e-2016-662 sin(bx) + bez poslba)). ~ y'= 49,64% +2% (663-262) cos(2x) * (662 +2 (3) sin(x)). Again differentiate worot "x"

=>" = Ibce y're fysieht te [1663-260)665lbs) – 1642+243) since) 2e 2 (EC3-2 (2) coolta) - 162 +263) sin(6x)) = e 2 - 6 (663-262) sin(a)_b/bc2 +263) Coslbx)). 42 -ae from initial conditions YLO)=9 - G+(2=9 y lo=42 = 4G+63-222 =24 => 24-C2+362512 y" (o)=-240 = 164 -2 1663-262) = 6L6C2 +263) = -240 => 164-12 gt4(2-3662-12435-240 - 164-322₂-24G = -240 => 26,-462-3₂ F-30 solving and ; 2G - C2+36==12 26-462-36=230 Z 46-5(2=-18 soling and 441+422=36 44 - 5(2= -18 962 = 54 [C2F6 As c2=6 then 623 substitute (=3, C2 = b then b-6+363 = 12 ino. Then substitute (83,22=b and (3=4

y=3040 -21 16 Coslba) +4 sin bx)) Answer ya 3e4a te 26cos(ba) + 4 sinal) sindow)