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)`,


`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)`.