Limit of a separable equation:Consider the equationdydt=k(a−y)(b−y)where a,b and k are constants. Assuming y(0)=0a) Solve for y(t) when a=bb) Solve for the case 0&lt;a&lt;bc) By considering the limit b→a in (b) show that the two results are consistent

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Limit of a separable equation:Consider the equationdydt=k(a−y)(b−y)where a,b and k are constants. Assuming y(0)=0a) Solve for y(t) when a=bb) Solve for the case 0&amp;lt;a&amp;lt;bc) By considering the limit b→a in (b) show that the two results are consistent

rebecosasny8a · Accepted Answer

Step 1The second solution does not go to 0. Let&#039;s call δ=b−a, and we&#039;ll replace b by a+δ. Also note that we can expand the exponential:exp⁡(δkt)≈1+δktStep 2Then y≈a(a+δ)(1+δkt−1)(a+δ)(1+δkt)−a=a2δkt+aδ2kta+aδkt+δ+δ2kt−aIgnore terms in δ2, and simplify by dividing both numerator and denominator by δ, and you get the answer you need.

Limit of a separable equation: Consider the equation \(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}}={k}{\left({a}-{y}\right)}{\left({b}-{y}\right)}\) where

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