[Math] Bernoulli Equation for First-Order Differential Equations

Form

$$ y' = a(t)y + f(t)y^n; n \neq 0, 1 $$

Solving Steps

$$
\begin{split}

y' &= a(t)y + f(t)y^n \\
\\
\text{let } z &= y^{1-n} \\
\dfrac{dz}{dt} &= \dfrac{dz}{dy} \dfrac{dy}{dt} \\
\dfrac{dz}{dt} &= \left ( y^{1-n} \right ) ' \cdot y' \\
\dfrac{dz}{dt} &= ( 1-n ) y^{-n} \cdot y' \\
\\
\dfrac{dz}{dt} &= (1-n) y^{-n} \cdot \left ( a(t) y + f(t) y^n \right ) \\
\dfrac{dz}{dt} &= (1-n) a(t) y^{1-n} + (1-n) f(t) \\
z' &= (1-n)a(t)z + (1-n) f(t)

\end{split}
$$

And now it's linear.

Example

Ex. 1

$$
\begin{split}

t^2 y' + 2 t y - y^3 &= 0 \\
t^2 \cdot y' &= -2 t y + y^3 \\
y' &= \dfrac{-2}{t} y + \dfrac{1}{t^2} y^3 \\
\\
a(t) &= \dfrac{-2}{t} \\
f(t) &= \dfrac{1}{t^2} \\
n &= 3 \\
\\
z' &= (1-3) \dfrac{-2}{t} z + \dfrac{1}{t^2} (1-3) \\
z' &= \dfrac{4}{t} z + \dfrac{-2}{t^2} \\
z' - \dfrac{4}{t} z &= \dfrac{-2}{t^2} \\
\\
\text{Integrating factor: } I &= e^{\int \frac{-4}{t} dt} \\
&= e^{-4 \ln \mid t \mid } \\
&= (e^{\ln \mid t \mid})^{-4} \\
&= \mid t \mid^{-4} \\
&= \dfrac{1}{ \mid t \mid^4 } \\
&= \dfrac{1}{t^4} \\
\\
\dfrac{1}{t^4} \left ( z' - \dfrac{4}{t} z \right ) &=
 \dfrac{1}{t^4} \left ( \dfrac{-2}{t^2} \right ) \\
\dfrac{1}{t^4} z' - \dfrac{4}{t^5} z &=
 \dfrac{-2}{t^6} \\
\left ( \dfrac{1}{t^4} z \right )' &= \dfrac{-2}{t^6} \\
\int \left ( \dfrac{1}{t^4} z \right ) ' dt &= \int \dfrac{-2}{t^6} dt \\
\dfrac{1}{t^4} z &= \dfrac{2}{5 t^5} + C \\
z &= \dfrac{2}{5t} + C t^4 \\
\\
z &= y^{1-n} = y^{-2} \\
\\
y^{-2} &= \dfrac{2}{5t} + C t^4 \\
\dfrac{1}{y^2} &= \dfrac{2 + 5 C t^5}{5t} \\
y^2 &= \dfrac{5t}{2 + 5 C t^5} \\
y &= \pm \sqrt{\dfrac{5t}{2+5Ct^5}}

\end{split}
$$

Reference

https://www.youtube.com/watch?v=hNCE3AxbWj0
https://www.youtube.com/watch?v=TnVq-JFJVHA

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