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emma29 Group Title

given (1+t)y' +y= cos(t) , y(0)=1 , I tried solving it using P(x) and q(x). So far I have d/dt(yt) = integral (cost/(1+t))*t dt...I'm stuck with the right side.

  • 2 years ago
  • 2 years ago

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  1. Algebraic! Group Title
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    think you're making it too hard...

    • 2 years ago
  2. Algebraic! Group Title
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    \[(t+1)y' + y = ( (t+1)*y)'\]

    • 2 years ago
  3. Algebraic! Group Title
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    make sense?

    • 2 years ago
  4. Algebraic! Group Title
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    hello?

    • 2 years ago
  5. emma29 Group Title
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    sorry i'm just trying it out

    • 2 years ago
  6. Algebraic! Group Title
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    k :)

    • 2 years ago
  7. emma29 Group Title
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    am i suppose to end up with something like this?\[y dy=cost/(t+1) dt\] and then integrate both sides?

    • 2 years ago
  8. Algebraic! Group Title
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    hmm you're mixing methods:) you have:

    • 2 years ago
  9. Algebraic! Group Title
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    \[( (t+1)*y)' = \cos(t)\]

    • 2 years ago
  10. Algebraic! Group Title
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    \[\int\limits_{ }^{ }( (t+1)*y)' = \int\limits_{ }^{ }\cos(t)\]

    • 2 years ago
  11. Algebraic! Group Title
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    \[(t+1)*y= \sin(t) +C\] whoops

    • 2 years ago
  12. emma29 Group Title
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    ok i'll try this out..thanks

    • 2 years ago
  13. Callisto Group Title
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    I'm too stupid to solve in this way... \[(1+t)y' +y= cos(t)\]\[y' +\frac{1}{1+t}y= \frac{1}{1+t}cos(t)\] Nice :) \[\alpha = e^{\int\frac{1}{1+t}dt}=e^{\ln|1+t|} = 1+t\]\[y=\frac{1}{\alpha}\int \alpha \times \frac{1}{1+t}cos(t) dt = \frac{1}{1+t}\int (1+t) \times \frac{1}{1+t}cos(t) dt \]\[= \frac{1}{1+t}\int cost dt =...\] Why can't I make my life easier :'(

    • 2 years ago
  14. emma29 Group Title
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    oh wow i made a mistake when taking the integral of \[\int\limits1/(1+t) \] no wonder it wouldn't come out nicely..thanks

    • 2 years ago
  15. emma29 Group Title
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    thanks to both for your help!!

    • 2 years ago
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spraguer (Moderator)
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is replying to Can someone tell me what button the professor is hitting...

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