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Solving Bernoulli Differential Equations

fact
A Bernoulli Differntial Equation is one with the form: y+p(x)y=q(x)yn
fact
We can solve a Bernoulli differential equation by using the substitution: v=y1n Which gives us the equation: 11nv+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)yn by yn: yyn+p(x)y1n=q(x) Use the substitution v=y1nv=(1n)yny (the derivative uses implicit differentiation which you should look up if you've forgotten) Now we can rewrite our equation as: 11nv+p(x)v=q(x) just as we wanted.
example

Solve the IVP: y+4xy=x3y2,y(2)=1

So p(x)=4x, q(x)=x3, n=2. Which means v=y1 and v=y2y Plugging into our formula we have: v+4xv=x3v4xv=x3 This is a linear differential equation just like the ones we've seen before so: μ=e4xdx=x4 v=x4(x4x3dx+c)=x4(x1dx+c) v=x4(cln(x)) We're almost there. Now that we have v we can plug it back into our original substitution v=y1n and find y. y1=x4(cln(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(cln(2)) 1=16c16ln(2) c=16ln(2)116=ln(1)116 And finally we can plug this into our equation for y and find that: y=1x4(ln(2)116ln(x))=16x4(16ln(2)+1+16ln(x))=16x4(1+16ln(x2))
practice problems
Solve \(y' + 39y = 32y^{-2}\)
(use \(c\) for unknown constant)

\(y(t) = \)

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