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

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amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0what methods do you know of?

haganmc
 3 years ago
Best ResponseYou've already chosen the best response.0undetermined coefficients

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0got a lousy connection i believe

Algebraic!
 3 years ago
Best ResponseYou've already chosen the best response.1happening to me too, OS is bugged today.

JakeV8
 3 years ago
Best ResponseYou've already chosen the best response.0site is slow for me too :(

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0well, 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 \[e^{t/2}x= \int e^{t/2}\frac{3t^2}{2}dt\]

Algebraic!
 3 years ago
Best ResponseYou've already chosen the best response.1trial 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

amistre64
 3 years ago
Best 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) \[2*\frac32e^{t/2}(t^24t+8)\]

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0forgot 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} \[x=3(t^24t+8)+Ce^{t/2}\]

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0my idea was undetermined coeffs tho was it

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0*wasnt .... ugh, when things slow down, they sloowww down

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