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Bernoulli Equations

A Bernoulli equation has the form y′+p(t)y=q(t)yn{y}'+{p}{\left({t}\right)}{y}={q}{\left({t}\right)}{{y}}^{{n}} where n{n} is a real number.

Using the substituion z=y1−n{z}={{y}}^{{{1}-{n}}}, this equation can be transformed into a linear one.

Example 1. Solve y′−3ty=t4y13{y}'-\frac{{3}}{{t}}{y}={{t}}^{{4}}{{y}}^{{\frac{{1}}{{3}}}}.

This is a Bernoulli equation. Use the substitution z=y1−13=y23{z}={{y}}^{{{1}-\frac{{1}}{{3}}}}={{y}}^{{\frac{{2}}{{3}}}}. Then, y=z32{y}={{z}}^{{\frac{{3}}{{2}}}}, and y′=32z12z′{y}'=\frac{{3}}{{2}}{{z}}^{{\frac{{1}}{{2}}}}{z}'. Plugging these values into the equation gives

32z12z′−3tz32=t4(z32)13\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′−2tz=23t4{z}'-\frac{{2}}{{t}}{z}=\frac{{2}}{{3}}{{t}}^{{4}}.

This is a linear equation.

The integrating factor is I=e∫−2tdt=e−2ln⁡(t)=1t2{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{I}: 1t2z′−2t3z=23t2\frac{{1}}{{{t}}^{{2}}}{z}'-\frac{{2}}{{{t}}^{{3}}}{z}=\frac{{2}}{{3}}{{t}}^{{2}}, or d(zt2)dt=23t2\frac{{{d}{\left(\frac{{z}}{{{t}}^{{2}}}\right)}}}{{{d}{t}}}=\frac{{2}}{{3}}{{t}}^{{2}}.

Integrating the last equation gives: zt2=29t3+C\frac{{z}}{{{t}}^{{2}}}=\frac{{2}}{{9}}{{t}}^{{3}}+{C}, or z=Ct2+29t5{z}={C}{{t}}^{{2}}+\frac{{2}}{{9}}{{t}}^{{5}}.

Recall that z=y23{z}={{y}}^{{\frac{{2}}{{3}}}}; so, the final solution in the implicit form is y23=Ct2+29t5{{y}}^{{\frac{{2}}{{3}}}}={C}{{t}}^{{2}}+\frac{{2}}{{9}}{{t}}^{{5}}.

Let's do some more practice with another example.

Example 2. Solve y2y′+y3=1{{y}}^{{2}}{y}'+{{y}}^{{3}}={1}, y(0)=2{y}{\left({0}\right)}={2}.

This equation is not in the standard form; so, divide both sides by y2{{y}}^{{2}}: y′+y=1y2{y}'+{y}=\frac{{1}}{{{y}}^{{2}}}. This is a Bernoulli equation (note that there is no function of t{t} on the right side).

Use the substitution z=y1−(−2)=y3{z}={{y}}^{{{1}-{\left(-{2}\right)}}}={{y}}^{{3}}; then, y=z13{y}={{z}}^{{\frac{{1}}{{3}}}}, and y′=13z−23z′{y}'=\frac{{1}}{{3}}{{z}}^{{-\frac{{2}}{{3}}}}{z}'. Plugging these values into the equation gives:

13z−23z′+z13=1(z13)2\frac{{1}}{{3}}{{z}}^{{-\frac{{2}}{{3}}}}{z}'+{{z}}^{{\frac{{1}}{{3}}}}=\frac{{1}}{{{\left({{z}}^{{\frac{{1}}{{3}}}}\right)}}^{{2}}}, or z′+3z=3{z}'+{3}{z}={3}.

This equation is linear. The integrating factor is I=e∫3dt=e3t{I}={{e}}^{{\int{3}{d}{t}}}={{e}}^{{{3}{t}}}.

Multiply both sides of the equation by I{I}: e3tz′+3e3tz=3e3t{{e}}^{{{3}{t}}}{z}'+{3}{{e}}^{{{3}{t}}}{z}={3}{{e}}^{{{3}{t}}}, or d(e3tz)dt=3e3t\frac{{{d}{\left({{e}}^{{{3}{t}}}{z}\right)}}}{{{d}{t}}}={3}{{e}}^{{{3}{t}}}.

Integrating both sides, we obtain that e3tz=e3t+C{{e}}^{{{3}{t}}}{z}={{e}}^{{{3}{t}}}+{C}, or z=Ce−3t+1{z}={C}{{e}}^{{-{3}{t}}}+{1}.

Since z=y3{z}={{y}}^{{3}}, we have that y3=Ce−3t+1{{y}}^{{3}}={C}{{e}}^{{-{3}{t}}}+{1}, or y=Ce−3t+13{y}={\sqrt[{{3}}]{{{C}{{e}}^{{-{3}{t}}}+{1}}}}.

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

y(0)=2=Ce−3⋅0+13{y}{\left({0}\right)}={2}={\sqrt[{{3}}]{{{C}{{e}}^{{-{3}\cdot{0}}}+{1}}}}, or C=7{C}={7}.

So, y=7e−3t+13{y}={\sqrt[{{3}}]{{{7}{{e}}^{{-{3}{t}}}+{1}}}}.