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Let $$\frac{dx}{dt}=\frac{1}{16}x(4-x)^2.$$ Solve for $x$ implicitly when $x(0)=3$.

Can anyone explain what this means or how to solve it. Thank you.

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Implicitly, you have already described what $x$ is. I think you seek an explicit solution. If you separate your variables, you get $$\frac{16\,dx}{x(4-x)^2}=dt$$ Partial fraction decomposition gives (steps left as a learning exercise) $$\frac{16}{x(4-x)^2}=- \frac{1}{x - 4} + \frac{4}{\left(x - 4\right)^{2}} + \frac{1}{x}$$ Then integrating these terms with respect to $x$ is simple (just $u$-sub with power rule and $\ln$), and integrating $dt$ is trivial. Solve for the constant of integration by subbing 0 for $t$ and 3 for $x$ since $x(0)=3$. I leave the details for you to do as a learning exercise.

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  • $\begingroup$ Thank you so much, it makes so much more sense now. $\endgroup$ – MANONMARS45 Oct 10 '18 at 23:42
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The complete solution for $$\frac{dx}{dt}=\frac{1}{16}x(4-x)^2$$ with $x(0) = 3$ is seen by the following.

Using $$\frac{16}{x (x-4)^2} = \frac{4}{(x-4)^2} - \frac{1}{x-4} + \frac{1}{x} = \frac{d}{dx} \left( 1 + \ln\left(\frac{x}{x-4} \, e^{-x/(x-4)} \right) \right)$$ and $y \, e^{y} = q$ has the Lambert W-function solution $y = W(q)$, then:

\begin{align} \frac{dx}{dt} &= \frac{x(x-4)^2}{16} \\ \frac{16 \, dx}{x(x-4)^2} &= dt \\ \int \frac{16 \, dx}{x(x-4)^2} &= t + c_{0} \\ \ln\left(\frac{x}{x-4} \, e^{-x/(x-4)} \right) &= t - 1 + c_{0} \\ \frac{x}{x-4} \, e^{-x/(x-4)} &= c_{1} \, e^{t-1}. \end{align} Now use $x(0) = 3$ to obtain $c_{1} = - 3 \, e^{4}$ and \begin{align} - \frac{x}{x-4} \, e^{-x/(x-4)} &= 3 \, e^{t + 3} \\ - \frac{x}{x-4} &= W(3 \, e^{t+3}) \end{align} or $$x(t) = \frac{4 \, W(3 \, e^{t+3})}{1 + W(3 \, e^{t+3})}.$$

To check the boundary condition use $W(x \, e^{x}) = x$. This is used in the following way: $$x(0) = \frac{4 \, W(3 \, e^{3})}{1 + W(3 \, e^{3})} = \frac{4 \cdot 3}{1 + 3} = 3$$ as required.

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