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amistre64 Group TitleBest ResponseYou've already chosen the best response.0
what methods do you know of?
 one year ago

haganmc Group TitleBest ResponseYou've already chosen the best response.0
undetermined coefficients
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
got a lousy connection i believe
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
happening to me too, OS is bugged today.
 one year ago

JakeV8 Group TitleBest ResponseYou've already chosen the best response.0
site is slow for me too :(
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
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 \[e^{t/2}x= \int e^{t/2}\frac{3t^2}{2}dt\]
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
\[\int e^{t/2}\frac{3t^2}{2}dt\] e^(t/2) +t^2 2e^(t/2) 2t 4e^(t/2) +2 8e^(t/2) \[2*\frac32e^{t/2}(t^24t+8)\]
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
forgot the +C at the end:) this gives us \[e^{t/2}x=3e^{t/2}(t^24t+8)+C\] divide off the e^{t/2} \[x=3(t^24t+8)+Ce^{t/2}\]
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
my idea was undetermined coeffs tho was it
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
*wasnt .... ugh, when things slow down, they sloowww down
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
I did it the 'integrating factor' way too, just because I was curious... fair bit of work doing that integral..
 one year ago
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