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anonymous

  • 5 years ago

solve the initial value problem du/dt = (t+1)/sqrt(t); u(1) = 4

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  1. amistre64
    • 5 years ago
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    \[\int\limits_{} \frac{t+1}{\sqrt{t}} dt\]

  2. amistre64
    • 5 years ago
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    split the fraction in two and work each side

  3. amistre64
    • 5 years ago
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    {S} t.(t^-1/2) + t^(-1/2) dt {S} t^(1/2) + t^(-1/2) dt

  4. amistre64
    • 5 years ago
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    t^(3/2) t^(1/2) ----- + ------ + C 3/2 1/2

  5. amistre64
    • 5 years ago
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    when t = 1; this should equal 4 according to the initial condition

  6. amistre64
    • 5 years ago
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    2/3 + 2 +C = 4 8/3 +C = 4 C = 4 - 8/3 = -16/3 if i did it right

  7. amistre64
    • 5 years ago
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    \[\frac{2t^2}{3}+\frac{2\sqrt{t}}{1}-\frac{16}{3}\]

  8. anonymous
    • 5 years ago
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    ahh thanks a ton. i was WAY overthinking it. used to other profs who constantly used all kinds of tricks and stuff so i was looking for something way more complicated. thanks again.

  9. anonymous
    • 5 years ago
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    though solving for c = 4/3, but otherwise it's golden

  10. amistre64
    • 5 years ago
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    :) youre welcome

  11. amistre64
    • 5 years ago
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    yeah, the integrating i can do.... addition? nah lol

  12. anonymous
    • 5 years ago
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    haha it's cool. i'm usually the same way- just apparently brain dead tonight

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