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haganmc

  • 3 years ago

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

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  1. amistre64
    • 3 years ago
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    what methods do you know of?

  2. haganmc
    • 3 years ago
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    undetermined coefficients

  3. amistre64
    • 3 years ago
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    got a lousy connection i believe

  4. Algebraic!
    • 3 years ago
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    happening to me too, OS is bugged today.

  5. JakeV8
    • 3 years ago
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    site is slow for me too :(

  6. amistre64
    • 3 years ago
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    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 \[e^{t/2}x= \int e^{t/2}\frac{3t^2}{2}dt\]

  7. Algebraic!
    • 3 years ago
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    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

  8. amistre64
    • 3 years ago
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    \[\int e^{t/2}\frac{3t^2}{2}dt\] e^(t/2) +t^2 2e^(t/2) -2t 4e^(t/2) +2 8e^(t/2) \[2*\frac32e^{t/2}(t^2-4t+8)\]

  9. amistre64
    • 3 years ago
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    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} \[x=3(t^2-4t+8)+Ce^{-t/2}\]

  10. amistre64
    • 3 years ago
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    my idea was undetermined coeffs tho was it

  11. amistre64
    • 3 years ago
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    *wasnt .... ugh, when things slow down, they sloowww down

  12. Algebraic!
    • 3 years ago
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    I did it the 'integrating factor' way too, just because I was curious... fair bit of work doing that integral..

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