A Bernoulli equation has the form y′+p(t)y=q(t)yn where n is a real number.
Using the substituion z=y1−n, this equation can be transformed into a linear one.
Example 1. Solve y′−t3y=t4y31.
This is a Bernoulli equation. Use the substitution z=y1−31=y32. Then, y=z23, and y′=23z21z′. Plugging these values into the equation gives
23z21z′−t3z23=t4(z23)31,
Or
z′−t2z=32t4.
This is a linear equation.
The integrating factor is I=e∫−t2dt=e−2ln(t)=t21.
Multiply both sides of the equation by I: t21z′−t32z=32t2, or dtd(t2z)=32t2.
Integrating the last equation gives: t2z=92t3+C, or z=Ct2+92t5.
Recall that z=y32; so, the final solution in the implicit form is y32=Ct2+92t5.
Let's do some more practice with another example.
Example 2. Solve y2y′+y3=1, y(0)=2.
This equation is not in the standard form; so, divide both sides by y2: y′+y=y21. This is a Bernoulli equation (note that there is no function of t on the right side).
Use the substitution z=y1−(−2)=y3; then, y=z31, and y′=31z−32z′. Plugging these values into the equation gives:
31z−32z′+z31=(z31)21, or z′+3z=3.
This equation is linear. The integrating factor is I=e∫3dt=e3t.
Multiply both sides of the equation by I: e3tz′+3e3tz=3e3t, or dtd(e3tz)=3e3t.
Integrating both sides, we obtain that e3tz=e3t+C, or z=Ce−3t+1.
Since z=y3, we have that y3=Ce−3t+1, or y=3Ce−3t+1.
Now, plug the initial conditions to find the particular solution: