I Can Differential Equations with Quadratic Terms Be Solved Analytically?

[davidge]

How can we solve a differential equation of the form $$a(r)Y' = Y - Y^2$$ for $Y \equiv \phi(r)$? An equation of this form appears in https://en.wikipedia.org/wiki/Deriv...field_equations_to_find_A.28r.29_and_B.28r.29. If there were not the $Y^2$, that would be a easy to solve differential equation: just a first order linear differential equation. The problem is the $Y^2$ though.

 

[Ssnow]

You must use the separable variables method.

 

[Ssnow]

Note: In order to have a closed form for the solution $Y$ you must integrate $\int \frac{dr}{a(r)}$, so it depends by the expression of $a(r)$ ...
Ssnow

 

[fresh_42]

This is a Bernoulli differential equation, which is a special case of a Riccati equation.

 

[davidge]

$\$

Note: In order to have a closed form for the solution $Y$ you must integrate $\int \frac{dr}{a(r)}$, so it depends by the expression of $a(r)$ ...
Ssnow

You must use the separable variables method.

Just too easy! Thanks for the hint

$$ a(r) \frac{dy}{dr} = y - y^2 \\
\frac{dy}{y} + \frac{dy}{1-y} = \frac{dr}{a(r)} \\
\text{ln}y + c_1 - \text{ln}(1-y) + c_2 = \int \frac{1}{a(r)}\ dr \\
C \equiv c_1 + c_2 \\
\text{ln}\bigg(\frac{y}{1-y}\bigg) + C = \int \frac{1}{a(r)} \ dr \\
\frac{y}{1-y} = e^{\int \frac{1}{a(r)}\ dr - C} \\
y\bigg(1- e^{\int \frac{1}{a(r)}\ dr - C}\bigg) = e^{\int \frac{1}{a(r)}\ dr - C} \\
\Rightarrow y = \frac{e^{\int \frac{1}{a(r)}\ dr - C}}{\bigg(1- e^{\int \frac{1}{a(r)}\ dr - C}\bigg)}$$