Q help pls (1 Viewer)

Lith_30 · Mar 23, 2022

For part a)

We need to get the DE $v'=r(1-v)$ therefore we need an equation for $\frac{dv}{dx}$

using the chain rule we can say $\frac{dv}{dx}=\frac{dy}{dx}\times\frac{dv}{dy}$

the question already says that $\frac{dy}{dx}=ry(1-y)$

to find $\frac{dv}{dy}$ we can use the given identity $v=\frac{1}{y}\implies\frac{dv}{dy}=-\frac{1}{y^2}$

so we can say that $\frac{dv}{dx}=r\frac{1}{v}(1-\frac{1}{v})\times-v^2$ (convert everthing in terms of v)

Which eventually simplifies down to $v'=r(1-v)$

part b)
$\\\frac{dv}{dx}=r(1-v)\\\\\frac{dv}{1-v}=r\ dx\\\\\int{\frac{dv}{1-v}}=r\int{dx}\\\\-\ln(1-v)=rx+c\\\\\ln(1-v)=-rx+c\\\\1-v=e^{-rx+c}\\v=1-Ae^{-rx}\ \text{Where a}=e^c$

part c)
We just make everything in terms of y

$\\\frac{1}{y}=1-Ae^{-rx}\\\\y=\frac{1}{1-Ae^{-rx}}$

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Q help pls (1 Viewer)

Akshara Patil

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Lith_30

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