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FibonacciChick666 Group Title

Hey, Quick help Please: Transform the given IVP into an equivalent problem with the initial pt. at the origin. \(\frac{dy}{dt}=t^2+y^2\) (I know it's easy... I'm just having a brain fart)

  • 11 months ago
  • 11 months ago

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  1. abb0t Group Title
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    put it in the form \(y'-t^2=y^2\)

    • 11 months ago
  2. FibonacciChick666 Group Title
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    whoops forgot the IV \(y(1)=2\)

    • 11 months ago
  3. FibonacciChick666 Group Title
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    I need a little more please @abb0t ... It's not my ODE type of day

    • 11 months ago
  4. FibonacciChick666 Group Title
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    Hey @hero could you take a look at this for me please, it's just not clicking at the moment

    • 11 months ago
  5. FibonacciChick666 Group Title
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    @amistre64, @mathstudent, or @satellite Do you mind taking a look at this for me please? You come highly recommended

    • 11 months ago
  6. dumbcow Group Title
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    oh oops i was trying to solve the ODE...so you just want to change ODE to reflect the IV at (0,0) plug in original IV (1,2) to determine dy/dt dy/dt = 1^2 + 2^2 = 5 now plug in (0,0) and keep dy/dt = 5 5 = 0^2 +0^2 + C 5 = C dy/dt = t^2 +y^2 +5

    • 11 months ago
  7. FibonacciChick666 Group Title
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    oh.... that was easy thanks... I had y'= 5t and I'm like this is sooo wrong ok ty again

    • 11 months ago
  8. dumbcow Group Title
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    yw i hope you dont have to solve this one...its ugly no elementary solution

    • 11 months ago
  9. FibonacciChick666 Group Title
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    yea I know... I just have a massive headache and screwed this up like 6 times so I was sick of it lol but nah just change to IV =0 but I was thinking integrating factor maybe

    • 11 months ago
  10. abb0t Group Title
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    I see someone helped you already. I had to go sorry.

    • 11 months ago
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