Solving Bernoulli Differential Equations

fact
A Bernoulli Differntial Equation is one with the form: $$y' + p(x)y = q(x)y^n$$
fact
We can solve a Bernoulli differential equation by using the substitution: $$v = y^{1-n}$$ Which gives us the equation: $$\frac{1}{1-n}v' + p(x)v = q(x)$$ Which is a linear differential equation, which we know how to solve.
derivation
Begin by dividing $$y' + p(x)y = q(x)y^n$$ by $$y^n$$: $$y'y^{-n} + p(x)y^{1-n} = q(x)$$ Use the substitution $$v = y^{1-n} \implies v' = (1-n)y^{-n}y'$$ (the derivative uses implicit differentiation which you should look up if you've forgotten) Now we can rewrite our equation as: $$\frac{1}{1-n}v' + p(x)v = q(x)$$ just as we wanted.
example

Solve the IVP: $$y' + \frac{4}{x}y = x^3y^2,\quad y(2) = -1$$

So $$p(x) = \frac{4}{x}$$, $$q(x) = x^3$$, $$n = 2$$. Which means $$v = y^{-1}$$ and $$v' = -y^{-2}y'$$ Plugging into our formula we have: $$-v' + \frac{4}{x}v = x^3 \implies v' - \frac{4}{x}v = -x^3$$ This is a linear differential equation just like the ones we've seen before so: $$\mu = e^{-\frac{4}{x}dx} = x^{-4}$$ $$v = x^4(\int -x^{-4}x^3dx + c) = -x^4(\int x^{-1}dx + c)$$ $$v = x^4(c - \ln(x))$$ We're almost there. Now that we have $$v$$ we can plug it back into our original substitution $$v = y^{1-n}$$ and find $$y$$. $$y^{-1} = x^4(c - \ln(x))$$ Now we can plug in our initial condition to solve for $$c$$. We could completely solve for $$y$$ and then plug in our initial condition however it looks easier to do it here before everything becomes a big fraction with logarithms and powers and such. $$(-1)^{-1} = 16(c - \ln(2))$$ $$-1 = 16c - 16\ln(2)$$ $$c = \frac{16\ln(2) - 1}{16} = \ln(1) - \frac{1}{16}$$ And finally we can plug this into our equation for $$y$$ and find that: $$y = \frac{1}{x^4(\ln(2) - \frac{1}{16} - \ln(x))} = \frac{-16}{x^4(-16\ln(2) + 1 + 16\ln(x))} = \frac{-16}{x^4(1 + 16\ln(\frac{x}{2}))}$$
practice problems