# Bernoulli Equations

A Bernoulli equation has the form y'+p(t)y=q(t)y^n where n is a real number.

Using the substituion z=y^(1-n), this equation can be transformed into a linear one.

Example 1. Solve y'-3/ty=t^4y^(1/3).

This is a Bernoulli equation. Use the substitution z=y^(1-1/3)=y^(2/3). Then, y=z^(3/2), and y'=3/2 z^(1/2)z'. Plugging these values into the equation gives

3/2 z^(1/2)z'-3/t z^(3/2)=t^4 (z^(3/2))^(1/3),

Or

z'-2/t z=2/3 t^4.

This is a linear equation.

The integrating factor is I=e^(int -2/t dt)=e^(-2ln(t))=1/t^2.

Multiply both sides of equation by I: 1/t^2 z'-2/t^3 z=2/3 t^2 or (d(z/t^2))/(dt)=2/3 t^2.

Integrating the last equation gives: z/t^2=2/9t^3+C or z=Ct^2+2/9 t^5.

Recall that z=y^(2/3); so, the final solution in the implicit form is y^(2/3)=Ct^2+2/9t^5.

Let's do some more practice with another example.

Example 2. Solve y^2y'+y^3=1, y(0)=2.

This equation is not in the standard form; so, divide both sides by y^2: y'+y=1/y^2. This is a Bernoulli equation (note that there is no function of t on the right side).

Use the substitution z=y^(1-(-2))=y^3; then, y=z^(1/3), and y'=1/3 z^(-2/3)z'. Plugging these values into the equation gives:

1/3 z^(-2/3)z'+z^(1/3)=1/(z^(1/3))^2, or z'+3z=3.

This equation is linear. The integrating factor is I=e^(int 3dt)=e^(3t).

Multiply both sides of the equation by I: e^(3t)z'+3e^(3t)z=3e^(3t), or (d(e^(3t)z))/(dt)=3e^(3t).

Integrating both sides, we obtain that e^(3t)z=e^(3t)+C, or z=Ce^(-3t)+1.

Since z=y^3, we have that y^3=Ce^(-3t)+1, or y=root(3)(Ce^(-3t)+1).

Now, plug initial conditions to find the particular solution:

y(0)=2=root(3)(Ce^(-3*0)+1) or C=7.

So, y=root(3)(7e^(-3t)+1).