Question

Find the function\(f\)such that\(\;{f^\prime }(x) = f(x)(1 - f(x)){\rm{ and }}f(0) = \frac{1}{2}{\rm{. }}\)

Short answer

\(f(x) = \frac{{{e^x}}}{{{e^x} + 1}}\)

Step-by-step solution

Definition

Separable differential equations can be written in the form\(\frac{{dy}}{{dx}} = {\rm{ }}f\left( x \right){\rm{ }}g\left( y \right)\), where\(x,y\)are the variables and are explicitly separated from each other.

Separable differential equation

Let\(y = f(x) \Rightarrow \frac{{dy}}{{dx}} = {f^\prime }(x)\), and then equation can be rewritten as:

\(\begin{aligned}{}{f^\prime }(x) = f(x)(1 - f(x))\\\frac{{dy}}{{dx}} = y(1 - y)\\\frac{1}{{y(1 - y)}}dy = dx\end{aligned}\)

Integrate

Integrating both sides:

\(\begin{aligned}{}\int {\frac{1}{{y(1 - y)}}} dy = \int 1 dx + C\\\int {\frac{{1 - y + y}}{{y(1 - y)}}} dy = \int 1 dx + C\\\int {\left( {\frac{1}{y} + \frac{1}{{1 - y}}} \right)} dy = \int 1 dx + C\end{aligned}\)

Then we get:

\(\ln |y| - \ln |1 - y| = \frac{{{x^{0 + 1}}}}{{0 + 1}} + C\) Using formulas\(\int {{x^n}} dx = \frac{{{x^{n + 1}}}}{{n + 1}},n \ne - 1;\int {\frac{1}{x}} dx = \ln |x|\)

\(\ln \left| {\frac{y}{{y - 1}}} \right| = x + C\) Using formulas\(n|x| - \ln |y| = \ln \left| {\frac{x}{y}} \right|\)

Apply initial condition

Using initial condition\(y(0) = \frac{1}{2}\)that means\(x = 0,y = \frac{1}{2}\).

\(\begin{aligned}{}\ln \left| {\frac{{\frac{1}{2}}}{{1 - \frac{1}{2}}}} \right| = 0 + C\\\ln (1) = 0 + C\\C = 0\end{aligned}\)

Find solution

Substitute value\(C = 0\)and solve for\(y\):

\(\begin{aligned}{}\ln \left| {\frac{y}{{1 - y}}} \right| = x + 0\;\;\;{\rm{ Since }}C = 0\\\ln \left| {\frac{y}{{1 - y}}} \right| = x\\\left| {\frac{y}{{1 - y}}} \right| = {e^x}\\\frac{y}{{1 - y}} = \pm {e^x}\end{aligned}\)

Since, \(y(0) = \frac{1}{2} > 0\)we have

\(\frac{y}{{1 - y}} = {e^x}\)

\(\begin{aligned}{}\frac{{1 - y}}{y} = \frac{1}{{{e^x}}}\\\frac{1}{y} - 1 = \frac{1}{{{e^x}}}\\\frac{1}{y} = \frac{1}{{{e^x}}} + 1\\\frac{1}{y} = \frac{{{e^x} + 1}}{{{e^x}}}\end{aligned}\)

\(f(x) = \frac{{{e^x}}}{{{e^x} + 1}}\)