Question

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?

Answer 1

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.

Answer 2

Is this a compulsory step for all D operator questions, checking whether the general solution works.

Answer 3

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.