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anonymous

  • 4 years ago

does the IVP dy/dt=y*t^(1/2), y(1)=1 have a unique solution on an interval containing t=1?

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  1. amistre64
    • 4 years ago
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    looks seperable to me

  2. amistre64
    • 4 years ago
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    \[ln|y| =\frac{2}{3}t^{3/2}+C \]

  3. anonymous
    • 4 years ago
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    IVP=initial value problem. i have y=e^(2/3t^(3/2))+1-e^(2/3) but it should be y=e^(2/3(t^(3/2)-1))...not the same

  4. amistre64
    • 4 years ago
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    \[\large y =exp(\frac{2}{3}t^{3/2}+C)\] \[y =exp(\frac{2}{3}t^{3/2})*exp(C)\] \[y =C_1\ exp(\frac{2}{3}t^{3/2})\] \[1 =C_1\ exp(\frac{2}{3})\] \[exp(-\frac23) =C_1\] if i see it right

  5. anonymous
    • 4 years ago
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    you say seperable meaning not unique? cuz it should be unique...

  6. amistre64
    • 4 years ago
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    seperable meaning its easy to see if theres a solution :)

  7. mathmate
    • 4 years ago
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    The solution is unique within an interval if the function y and its derivative are continuous on the interval.

  8. amistre64
    • 4 years ago
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    \[\frac{dy}{dx}=f(x)\ h(y)\] \[\frac{1}{h(y)}dy=f(x)dx\] \[\int(\frac{1}{h(y)}dy=f(x)dx)\]

  9. anonymous
    • 4 years ago
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    i see where i messed up. but why do you have -2/3?^^^

  10. amistre64
    • 4 years ago
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    divide off the e^{2/3}

  11. amistre64
    • 4 years ago
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    division makes an exponent go negative

  12. anonymous
    • 4 years ago
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    is that 1/e^(2/3)?

  13. anonymous
    • 4 years ago
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    ok

  14. amistre64
    • 4 years ago
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    yes

  15. anonymous
    • 4 years ago
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    same question...\[dy/dt=6y ^{2/3}\] y(1)=0. i'm not getting 0 so i need verification. i got \[y=(2t-C)^{3}\] C=-8/19...

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