Solve (D^2-3D+2)y =x(x+4) and show that its general solution is given by y=Ae^x + Be^2x +(x^2/2) + (7x/2) + (19/4)

My work :

auxiliary eqn ,
m^2 - 3m + 2= 0
m=1 , m=2

Reduced eqn :

yn = Ae^x + Be^2x

P.I. = 1/f(D)*F(x) = (1/(D^2-3D+2))*(x(x+4))

P.I. = [(x^2+4x) + ( (1/2)*( 2- 3(2x) - 3(4) ) ) + (1/4)(9*2) )

P.I. = x^2 + (4x/2) - (6x/4) - (5/2) + (18/4)

P.I. =(x^2/2) + x + 8/4

General solution :

y= Ae^x + Be^2x + (x^2/2) + x + 8/4

Could anyone point out my mistakes?

3 answers

Well, your final mistake was not checking to make sure your general solution worked.

to find one solution, set
u = 1/(D-2) x(x+4)
(D-2) u = x^2+4x
Now use the integrating factor e^(-2x) to get
(e^(-2x) u)' = (x^2+4x) e^(-2x)
e^(-2x) u = ∫ (x^2+4x) e^(-2x) dx = -1/4 e^(-2x) (2x^2+10x+5)
u = -1/4 (2x^2+10x+5)

Now, y = 1/(D-1) u
follow that through and you wind up with
y = Ae^x + Be^2x + 1/4 (2x^2 + 14x + 19)

Or, you can decompose 1/f(D) using partial fractions.
Is this a compulsory step for all D operator questions, checking whether the general solution works.
it ought to be compulsory when solving any math problem!
The fact that you arrive at an answer does not mean it is correct. We all make mistakes.