Bernoulli Equations

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

Using substituion `z=y^(1-n)` this equation can be transformed into linear.

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

This is Bernoulli equation. Use 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 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 linear equation.

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 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 finally solution in implicit form is `y^(2/3)=Ct^2+2/9t^5` .

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

This equation is not in standard form, so divide both sides by `y^2` : `y'+y=1/y^2` . This is Bernoulli equation (note that there is not function of t on the right side).

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

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

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

Multiply both sides of 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` then `y^3=Ce^(-3t)+1` or `y=root(3)(Ce^(-3t)+1)`

Now, plug initial conditions to find particular solution:

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

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