Question

find particular solution to 2x' + x=3t^2

Answer 1

what methods do you know of?

Answer 2

undetermined coefficients

Answer 3

got a lousy connection i believe

Answer 4

happening to me too, OS is bugged today.

Answer 5

site is slow for me too :(

Answer 6

well, lets start by dividing off that 2

x' + x/2 = 3t^2/2

we want to determine an e^rt that will allow us to turn this into a product rule; e^t/2 should work

e^(t/2) x' +e^(t/2) x/2 = e^(t/2)3t^2/2

when we integrate both sides we get

\[e^{t/2}x= \int e^{t/2}\frac{3t^2}{2}dt\]

Answer 7

trial soln. is at^2 + bt +c

2x' = 4at + 2b
x= at^2 +bt +c
a=3
4a+b =0
2b+c =0

b=-12
c=24

Answer 8

\[\int e^{t/2}\frac{3t^2}{2}dt\]

e^(t/2)
+t^2 2e^(t/2)
-2t 4e^(t/2)
+2 8e^(t/2)

\[2*\frac32e^{t/2}(t^2-4t+8)\]

Answer 9

forgot the +C at the end:)

this gives us
\[e^{t/2}x=3e^{t/2}(t^2-4t+8)+C\]
divide off the e^{t/2}
\[x=3(t^2-4t+8)+Ce^{-t/2}\]

Answer 10

my idea was undetermined coeffs tho was it

Answer 11

*wasnt .... ugh, when things slow down, they sloowww down

Answer 12

I did it the 'integrating factor' way too, just because I was curious...
fair bit of work doing that integral..