# 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