Bernoulli Equations

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 $$${{y}}^{{\frac{{2}}{{3}}}}={C}{{t}}^{{2}}+\frac{{2}}{{9}}{{t}}^{{5}}$$$.

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