# Bernoulli Equations

A Bernoulli equation has the form ${y}'+{p}{\left({t}\right)}{y}={q}{\left({t}\right)}{{y}}^{{n}}$ 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 ${y}'-\frac{{3}}{{t}}{y}={{t}}^{{4}}{{y}}^{{\frac{{1}}{{3}}}}$.

This is a Bernoulli equation. Use the substitution ${z}={{y}}^{{{1}-\frac{{1}}{{3}}}}={{y}}^{{\frac{{2}}{{3}}}}$. Then, ${y}={{z}}^{{\frac{{3}}{{2}}}}$, and ${y}'=\frac{{3}}{{2}}{{z}}^{{\frac{{1}}{{2}}}}{z}'$. Plugging these values into the equation gives

$\frac{{3}}{{2}}{{z}}^{{\frac{{1}}{{2}}}}{z}'-\frac{{3}}{{t}}{{z}}^{{\frac{{3}}{{2}}}}={{t}}^{{4}}{{\left({{z}}^{{\frac{{3}}{{2}}}}\right)}}^{{\frac{{1}}{{3}}}}$,

Or

${z}'-\frac{{2}}{{t}}{z}=\frac{{2}}{{3}}{{t}}^{{4}}$.

This is a linear equation.

The integrating factor is ${I}={{e}}^{{\int-\frac{{2}}{{t}}{d}{t}}}={{e}}^{{-{2}{\ln{{\left({t}\right)}}}}}=\frac{{1}}{{{t}}^{{2}}}$.

Multiply both sides of the equation by ${I}$: $\frac{{1}}{{{t}}^{{2}}}{z}'-\frac{{2}}{{{t}}^{{3}}}{z}=\frac{{2}}{{3}}{{t}}^{{2}}$, or $\frac{{{d}{\left(\frac{{z}}{{{t}}^{{2}}}\right)}}}{{{d}{t}}}=\frac{{2}}{{3}}{{t}}^{{2}}$.

Integrating the last equation gives: $\frac{{z}}{{{t}}^{{2}}}=\frac{{2}}{{9}}{{t}}^{{3}}+{C}$, or ${z}={C}{{t}}^{{2}}+\frac{{2}}{{9}}{{t}}^{{5}}$.

Recall that ${z}={{y}}^{{\frac{{2}}{{3}}}}$; so, the final solution in the implicit form is ${{y}}^{{\frac{{2}}{{3}}}}={C}{{t}}^{{2}}+\frac{{2}}{{9}}{{t}}^{{5}}$.

Let's do some more practice with another example.

Example 2. Solve ${{y}}^{{2}}{y}'+{{y}}^{{3}}={1}$, ${y}{\left({0}\right)}={2}$.

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

Use the substitution ${z}={{y}}^{{{1}-{\left(-{2}\right)}}}={{y}}^{{3}}$; then, ${y}={{z}}^{{\frac{{1}}{{3}}}}$, and ${y}'=\frac{{1}}{{3}}{{z}}^{{-\frac{{2}}{{3}}}}{z}'$. Plugging these values into the equation gives:

$\frac{{1}}{{3}}{{z}}^{{-\frac{{2}}{{3}}}}{z}'+{{z}}^{{\frac{{1}}{{3}}}}=\frac{{1}}{{{\left({{z}}^{{\frac{{1}}{{3}}}}\right)}}^{{2}}}$, or ${z}'+{3}{z}={3}$.

This equation is linear. The integrating factor is ${I}={{e}}^{{\int{3}{d}{t}}}={{e}}^{{{3}{t}}}$.

Multiply both sides of the equation by ${I}$: ${{e}}^{{{3}{t}}}{z}'+{3}{{e}}^{{{3}{t}}}{z}={3}{{e}}^{{{3}{t}}}$, or $\frac{{{d}{\left({{e}}^{{{3}{t}}}{z}\right)}}}{{{d}{t}}}={3}{{e}}^{{{3}{t}}}$.

Integrating both sides, we obtain that ${{e}}^{{{3}{t}}}{z}={{e}}^{{{3}{t}}}+{C}$, or ${z}={C}{{e}}^{{-{3}{t}}}+{1}$.

Since ${z}={{y}}^{{3}}$, we have that ${{y}}^{{3}}={C}{{e}}^{{-{3}{t}}}+{1}$, or ${y}={\sqrt[{{3}}]{{{C}{{e}}^{{-{3}{t}}}+{1}}}}$.

Now, plug the initial conditions to find the particular solution:

${y}{\left({0}\right)}={2}={\sqrt[{{3}}]{{{C}{{e}}^{{-{3}\cdot{0}}}+{1}}}}$, or ${C}={7}$.

So, ${y}={\sqrt[{{3}}]{{{7}{{e}}^{{-{3}{t}}}+{1}}}}$.