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=y1−n
Which gives us the equation:
11−nv′+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:
y′y−n+p(x)y1−n=q(x)
Use the substitution v=y1−n⟹v′=(1−n)y−ny′ (the derivative uses implicit differentiation which you should look up if you've forgotten)
Now we can rewrite our equation as:
11−nv′+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=y−1 and v′=−y−2y′ Plugging into our formula we have: −v′+4xv=x3⟹v′−4xv=−x3 This is a linear differential equation just like the ones we've seen before so: μ=e−4xdx=x−4 v=x4(∫−x−4x3dx+c)=−x4(∫x−1dx+c) v=x4(c−ln(x)) We're almost there. Now that we have v we can plug it back into our original substitution v=y1−n and find y. y−1=x4(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−16ln(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)−116−ln(x))=−16x4(−16ln(2)+1+16ln(x))=−16x4(1+16ln(x2))
practice problems