Solving Bernoulli Differential Equations

A Bernoulli Differntial Equation is one with the form: $$y' + p(x)y = q(x)y^n$$

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.

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.

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: $y = \frac{1}{x^4(\ln(2) - \frac{1}{16} - \ln(x))} = \frac{-16}{x^4(-16\ln(2) + 1 + 16\ln(x))} = \frac{-16}{x^4(1 + 16\ln(\frac{x}{2}))}$