# 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) .