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anonymous

  • 5 years ago

Calculus Question (see next post for full question)

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  1. anonymous
    • 5 years ago
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    Every time I attempt to solve this question, the math gets complicated and frustrating. Can anyone help? Thanks! For many differential equations, the easiest way to find inflection points is to use the differential equation rather than the solution itself. To do this, we can compute y′′ by differentiating y′, remembering to use the chain rule wherever y occurs. Next, we can substitute for y′ by using the differential equation and setting y"=0. Then we can solve for y to find the inflection points. (Keep in mind here that solving for y can also produce some equilibrium solutions, which may not be inflection points!) Use the technique described above to find the inflection point for the solutions of the differential equation y′=r( 1- (y/L) )y Your answer may contain r and L.

  2. anonymous
    • 5 years ago
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    y' = ry - ry^2/L

  3. anonymous
    • 5 years ago
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    y" = ry' - 2ryy'/L

  4. anonymous
    • 5 years ago
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    0 = y'(r - 2ry/L) 0 = (ry - ry^2/L)(r - 2ry/L) Zero property, solve each parenth for y

  5. anonymous
    • 5 years ago
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    ry - ry^2/L = 0 --> ry(1-y/L) = 0 --> y = 0 and y = L r - 2ry/L = 0 --> 1 = 2y/L --> y = L/2

  6. anonymous
    • 5 years ago
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    I didn't check whether any are equilibrium points, but I hope my work helped!

  7. anonymous
    • 5 years ago
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    Thank you so much! The answer was correct. I'm just gonna see how my work matches yours now. :) I should be able to get the rest of the problems now. Thanks! Cheers.

  8. anonymous
    • 5 years ago
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    What exactly is the zero property?

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