Question

Need some help please..
Obtain the general solution:
(D^2 +3D +2)y = 12x^2

Answer 1

Auxiliary equation is
D^2+3D+2=0
D^2+2D+D+2=0
D(D+2)+1(D+2)=0
(D+2)(D+1)=0
D=-1,-2
\[General~ solution =C1e ^{-x}+C2e ^{-2x}\]

Answer 2

Right, I can get that far but the total answer is:
\[y=C1e^{-x} + C2e^{-2x} + 6x^2 -18x +2\]
I need help on the 2nd part of the answer please..

Answer 3

to find particular integral
let\[ y=ax^2+bx+c\]
\[D=\frac{ dy }{ dx }=2ax+b\]
\[D^2=\frac{ d^2D }{ dx^2 }=2a\]

Answer 4

put in the give eq.and find a,b,c
i am leaving.

Answer 5

So is it..
\[2A +3(2Ax+B)+2(Ax^2+Bx+C)=12x^2\] ?

Answer 6

The answer i gave, the last number should be a 21

Answer 7

\[2a+6ax+3b+2(ax^2+bx+c)=12x^2\]
\[2ax^2+(2b+6a)x+2a+3b+2c=12x^2\]
2a=12
a=6
2b+6*6=0
\[b=\frac{ -36 }{2 }=-18\]
2a+3b+2c=0
2*6+3*-18+2c=0
2c=54-12=42
\[c=\frac{ 42 }{2 }=21\]
\[hence P.I.=6x^2-18x+21\]
\[c.s.~ is~y=C1e ^{-x}+C2e ^{-2x}+6x^2-18x+21\]

Answer 8

You are awesome! Thank you!

Answer 9

YW