need help finding the gen. sol. of the differential equation (D-1)^2 (D+3)(D^2 + 2D + 5)^2[y] = 0

Question

lgbasallote · Answer

this is $$\huge [(D-1)^2 (d+3)(D^2 + 2D +5)^2] y = 0$$
right?

lgbasallote · Answer

interesting

lgbasallote · Answer

so what would be the problem?

anonymous · Answer

how would i turn those back into y-primes?

lgbasallote · Answer

why would you want to turn them into y primes?

anonymous · Answer

good question. i shouldnt

anonymous · Answer

in the form of x in a general solution

lgbasallote · Answer

have you forgotten that D = m?

you can just substitute straight to m

$$\huge [(m-1)^2 ( m+3)(m^2 + 2m + 5)^2]y = 0$$

lgbasallote · Answer

can you see it now?

anonymous · Answer

i see.  I will try to get it.  I just dont like the looks of two of them being squared

lgbasallote · Answer

haha it's no worries...it just means the root has multiplicity 2

anonymous · Answer

would you mind showing me how it is done for the (m-1)^2?  That will give me a great base for the long square

lgbasallote · Answer

i'll just give an example

suppose you have $$\huge (m+3)^2 (m-4) = 0$$
the roots are m = -3 ( 2x) and m=4

so the general solution is $$\huge y = y_c = c_1 e^{-3x} + c_2 x e^{-3x} + c_3 e^{4x}$$

lgbasallote · Answer

did that help?

anonymous · Answer

daaaamn this is a long problem

anonymous · Answer

yes. thank you again

lgbasallote · Answer

hahaha welcome ^_^

lgbasallote · Answer

and it's not really a long problem...just do q.f. in that long trinomial

anonymous · Answer