Answer:
[tex]y=\frac{e^{2t}}{9+e^{2t}}[/tex]
Step-by-step explanation:
We are given that
[tex]\frac{dy}{dt}=2y(1-y)[/tex]
[tex]y(0)=10%=\frac{10}{100}=0.1[/tex]
[tex]\frac{dy}{y(1-y)}=2dt[/tex]
[tex]\int\frac{dy}{y(1-y)}=\int 2dt[/tex]
[tex]\int(\frac{1}{y}+\frac{1}{1-y})dt=2\int dt[/tex]
[tex]ln y-ln(1-y)=2t+C[/tex]
By using the formula
[tex]\int \frac{dx}{x}=ln x[/tex]
[tex]ln\frac{y}{1-y}=2t+C[/tex]
By using the formula
[tex]ln m-ln n=ln(\frac{m}{n})[/tex]
[tex]\frac{y}{1-y}=e^{2t+C}=e^{2t}\cdot e^C=ce^{2t}[/tex]
Where [tex]e^C=c[/tex]
Substitute the values
[tex]\frac{0.1}{1-0.1}=c[/tex]
[tex]\frac{0.1}{0.9}=c[/tex]
[tex]c=\frac{1}{9}[/tex]
Substitute the value
[tex]\frac{y}{1-y}=\frac{1}{9}e^{2t}[/tex]
[tex]9y=e^{2t}-ye^{2t}[/tex]
[tex]9y+ye^{2t}=e^{2t}[/tex]
[tex]y(9+e^{2t})=e^{2t}[/tex]
[tex]y=\frac{e^{2t}}{9+e^{2t}}[/tex]
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The equation is [tex]y=\frac{9e^{2t}}{1+9e^{2t}}[/tex]
Given that: [tex]\frac{dy}{dt}=2y(1-y)[/tex]
We first solve this differential equation.
[tex]\frac{dy}{dt}=2y(1-y)\\\frac{dy}{y\left(1-y\right)}=2dt\\\frac{\left(y+\left(1-y\right)\right)}{y\left(1-y\right)}dy=2dt\\\left(\frac{1}{1-y}+\frac{1}{y}\right)dy=2dt[/tex]
Integrating both sides we get:
[tex]\int_{ }^{ }\left(\frac{1}{1-y}+\frac{1}{y}\right)dy=\int_{ }^{ }2dt\\-\ln\left(1-y\right)+\ln\left(y\right)=2t+c\\\ln\left(\frac{y}{1-y}\right)=2t+c\\\frac{y}{1-y}=e^{\left(2t+c\right)}\\\\\frac{y}{1-y}=e^{2t}\cdot e^{c}[/tex]
Let, [tex]C=e^{c}[/tex]
At t=0, y=10%=0.10
Using this, C = 9.
So the equation is:
[tex]\frac{y}{1-y}=9e^{2t}\\y=9e^{2t}-9e^{2t}y\\y\left(1+9e^{2t}\right)=9e^{2t}\\y=\frac{9e^{2t}}{1+9e^{2t}}[/tex]
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