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megannicole51

  • 2 years ago

solve the differential equation. assume x,y,t>0 dy/dt=-yln(y/2). y(0)=1

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  1. megannicole51
    • 2 years ago
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    so i think im doing this right so far but I'm not really sure can someone work through the steps with me?

  2. ddog1437
    • 2 years ago
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    are you taking calculus though?

  3. megannicole51
    • 2 years ago
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    calc 2

  4. megannicole51
    • 2 years ago
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    \[\frac{ dy }{ -ylny } = \frac{ 1 }{ 2 }dt\]

  5. megannicole51
    • 2 years ago
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    ill show the steps i have so far and can u tell me if im right or not?

  6. megannicole51
    • 2 years ago
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    ive never seen it done like that actually....my professor hasnt really done any examples so i am trying to figure it out from our book which isnt helping either

  7. megannicole51
    • 2 years ago
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    @SithsAndGiggles

  8. myininaya
    • 2 years ago
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    \[\frac{-1}{y \ln (\frac{y}{2})} dy= dt \]

  9. myininaya
    • 2 years ago
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    He is doing separation of variables.

  10. myininaya
    • 2 years ago
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    I think he meant to divide -yln(y/2) on both sides.

  11. myininaya
    • 2 years ago
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    and then he multiplied both sides by dt

  12. SithsAndGiggles
    • 2 years ago
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    Thanks to @myininaya for pointing out my mistake: \(\color{blue}{\text{Originally Posted by}}\) @SithsAndGiggles \[\frac{dy}{dt}=-y\ln\left(\frac{y}{2}\right)\\ -\frac{1}{y\ln\left(\frac{y}{2}\right)}~dy=dt\] Integrate both sides: \[-\int\frac{1}{y\ln\left(\frac{y}{2}\right)}~dy=\int dt\] First, a substitution: \(u=\dfrac{y}{2}\), or \(2u=y\), so that \(2du=dy\): \[-\int\frac{1}{2u\ln u}~(2~du)=\int dt\\ -\int\frac{1}{u \ln u}~du=\int dt\] Got everything so far? \(\color{blue}{\text{End of Quote}}\)

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