Bernoulli Equations, differential equations lesson

The logistic equation $y' = ry(1 - y/K)$ , with time $t$ as the independent variable and $y' = dy/dt$ , would be linear without its $y^2$ term. Equations that are linear except for one power of the unknown on the right side are Bernoulli equations. Dividing by that power and taking a suitable power of $y$ as the new unknown turns them into linear equations, which an integrating factor solves. This lesson derives the substitution, works an example through to a checked answer, and shows which solutions the substitution cannot see.

Key terms

Bernoulli equation

Definition With $t$ as the independent variable and $y' = dy/dt$ , a Bernoulli equation has the form

y' + p(t)\,y = q(t)\,y^n,

where $p$ and $q$ are continuous on an open interval $I$ and $n$ is a real constant. The coefficient of $y'$ must be $1$ , so an equation $a(t)y' + b(t)y = g(t)y^n$ is first divided by $a(t)$ on an interval where $a(t) \ne 0$ . For $n = 0$ and $n = 1$ the equation is linear, and for every other $n$ it is nonlinear. The power must make sense for the values of $y$ involved. A positive integer $n$ allows every real $y$ , a negative integer $n$ excludes $y = 0$ , and for a non-integer $n$ the usual real power requires $y \ge 0$ , or $y > 0$ when $n < 0$ .

In plain words A Bernoulli equation looks like a linear equation, except that its right side is multiplied by a power of the unknown. That power is the only nonlinear part.

Example The logistic equation $y' = ry(1 - y/K)$ with constants $r > 0$ and $K > 0$ can be written as

y' - ry = -\frac{r}{K}\,y^2,

so it is a Bernoulli equation with $n = 2$ , $p(t) = -r$ and $q(t) = -r/K$ . By contrast, $y' = 1 + y^2$ is not a Bernoulli equation, because its right side mixes two different powers of $y$ and only a first power can be moved to the left side.

Power substitution

Definition For $n \ne 0, 1$ , the power substitution replaces the unknown $y$ by

z = y^{1-n}.

On an interval where $y \ne 0$ , and $y > 0$ if $n$ is not an integer, the chain rule gives

z' = (1 - n)\,y^{-n}\,y'.

The exponent $1 - n$ is chosen because dividing the Bernoulli equation by $y^n$ gives $y^{-n}y' + p(t)\,y^{1-n} = q(t)$ . In this equation $y^{1-n}$ appears, and $y^{-n}y'$ is its derivative up to the constant factor $1 - n$ .

In plain words Divide by the power that causes the trouble and look at what remains. The term $p(t)\,y^{1-n}$ suggests the new unknown, and the chain rule shows that its derivative is already in the equation.

Example For $y' - y = -y^3$ we have $n = 3$ , so $z = y^{-2}$ and $z' = -2y^{-3}y'$ . Multiplying the equation by $-2y^{-3}$ gives

-2y^{-3}y' + 2y^{-2} = 2,

that is $z' + 2z = 2$ . Its solutions are $z = 1 + Ce^{-2t}$ , so $y^2 = 1/(1 + Ce^{-2t})$ . The new unknown $z = y^{-2}$ forgets the sign of $y$ , so the sign has to come from the data. With $y(0) = -\tfrac{1}{2}$ we get $z(0) = 4$ , so $C = 3$ and

y = -\frac{1}{\sqrt{1 + 3e^{-2t}}},

which is defined for every real $t$ .

Reduction to a linear equation

Definition Let $n \ne 0, 1$ and let $p$ and $q$ be continuous on $I$ . On a subinterval where $y \ne 0$ , and $y > 0$ if $n$ is not an integer, a differentiable function $y$ solves the Bernoulli equation exactly when $z = y^{1-n}$ solves the linear equation

z' + (1 - n)\,p\,z = (1 - n)\,q.

The reason is that the two equations differ by the factor $(1 - n)y^{-n}$ , which is never zero there. The linear equation has continuous coefficients, so it is solved with the integrating factor $e^{(1-n)\int p(t)\,dt}$ , and through any initial value its solution exists on all of $I$ . The solution $y$ is then recovered from $y^{1-n} = z$ , choosing the root or the sign that agrees with the initial value. It is valid only while $z$ stays nonzero, and positive when an even root or a non-integer power is needed to recover $y$ , so its interval can be shorter than $I$ . For $n > 1$ a zero of $z$ is a point where $y$ blows up.

In plain words Once the power has a new name, the equation is linear, and the integrating factor for linear equations solves it. The only new work is turning $z$ back into $y$ and checking where that is allowed.

Example Solve $y' + y = y^2$ with $y(0) = \tfrac{1}{2}$ . Here $n = 2$ and $z = 1/y$ , so $z' = -y'/y^2$ . Dividing the equation by $y^2$ and multiplying by $-1$ gives $z' - z = -1$ . With the integrating factor $e^{-t}$ ,

\begin{gathered} \left(e^{-t}z\right)' = -e^{-t}, \\[0.4em] e^{-t}z = e^{-t} + C, \end{gathered}

so $z = 1 + Ce^t$ . The initial value $z(0) = 2$ gives $C = 1$ , and

y = \frac{1}{1 + e^t}.

To check, $y' = -e^t/(1 + e^t)^2$ , and over the same denominator

\begin{aligned} y^2 - y &= \frac{1 - (1 + e^t)}{(1 + e^t)^2} \\[0.4em] &= -\frac{e^t}{(1 + e^t)^2} = y'. \end{aligned}

The denominator $1 + e^t$ is positive for every $t$ , so this solution exists on the whole line, with $0 < y < 1$ , and it decreases toward $0$ as $t \to \infty$ . Other members of the family $y = 1/(1 + Ce^t)$ behave differently. The value $C = 0$ gives the equilibrium $y = 1$ , and any $C < 0$ gives a pole where $1 + Ce^t = 0$ .

Exceptional exponents

Definition The exponents $n = 0$ and $n = 1$ are exceptional, because for them the Bernoulli equation is already linear and no substitution is needed. For $n = 0$ it reads $y' + p(t)y = q(t)$ . For $n = 1$ the two terms in $y$ combine into the homogeneous linear equation

y' + \bigl(p(t) - q(t)\bigr)y = 0.

For $n = 1$ the power substitution would even fail, since $z = y^0 = 1$ is constant and the factor $1 - n$ is $0$ . The other exponents are not exceptional, but some need care with the domain, as a negative $n$ excludes $y = 0$ and a non-integer $n$ restricts $y$ to nonnegative values.

In plain words When the power is $0$ or $1$ , there is nothing nonlinear to remove, so the equation is solved directly as a linear one.

Example The equation $y' + y = ty$ has $n = 1$ , $p(t) = 1$ and $q(t) = t$ . Moving the right side over gives $y' + (1 - t)y = 0$ , and the integrating factor $e^{t - t^2/2}$ gives the solutions

y = Ce^{t^2/2 - t}

for every real constant $C$ , on the whole line. The zero solution is the case $C = 0$ , because nothing was divided by $y$ . With $n = 0$ , as in $y' + y = t$ , the equation is linear from the start, with solutions $y = t - 1 + Ce^{-t}$ .

Zero solution

Definition For every $n > 0$ the zero function $y = 0$ solves the Bernoulli equation on all of $I$ , since both sides vanish. The power substitution assumes $y \ne 0$ , so this solution never comes out of the formula for $y$ and must be restored by checking the original equation. For $n = 0$ the zero function is a solution only when $q$ is identically zero, and for $n < 0$ it is not allowed at all. For an integer $n \ge 2$ the right side $q(t)y^n - p(t)y$ of $y' = q(t)y^n - p(t)y$ has a continuous partial derivative in $y$ , so solutions are unique, and a solution that is nonzero at one time is nonzero on its whole interval. For $0 < n < 1$ that partial derivative is unbounded near $y = 0$ wherever $q \ne 0$ , uniqueness can fail there, and a solution can stay at zero for a while and then leave it.

In plain words Dividing by $y^n$ quietly assumes $y \ne 0$ . Before dividing, check whether $y = 0$ solves the equation, and add it to the answer if it does.

Example For $y' + y = y^2$ , substituting $y = 0$ gives $0 = 0$ , so the zero function is a solution, but no constant $C$ in $y = 1/(1 + Ce^t)$ produces it. The complete set of solutions is that family together with $y = 0$ . For $y' = \sqrt{y}$ , a Bernoulli equation with $p = 0$ , $q = 1$ and $n = \tfrac{1}{2}$ , the initial value $y(0) = 0$ has the zero solution and also, for every $a \ge 0$ , the solution

y(t) = \begin{cases} 0, & t \le a, \\ \tfrac{1}{4}(t - a)^2, & t > a. \end{cases}

The substitution $z = \sqrt{y}$ gives $z' = \tfrac{1}{2}$ and only the pieces $y = \tfrac{1}{4}(t - c)^2$ with $t > c$ , so it cannot detect where a solution leaves zero.

Check yourself

Write $ty' + y = t^2y^3$ , for $t > 0$ , in Bernoulli form, and find the linear equation for the new unknown.

Show the answer

Dividing by $t$ gives the normalized form $y' + \frac{1}{t}\,y = t\,y^3$ , so $n = 3$ , $p(t) = 1/t$ and $q(t) = t$ . The new unknown is $z = y^{-2}$ , with $z' = -2y^{-3}y'$ . Multiplying the normalized equation by $-2y^{-3}$ gives

z' - \frac{2}{t}\,z = -2t,

valid on intervals with $t > 0$ and $y \ne 0$ . With the integrating factor $t^{-2}$ its solutions are $z = t^2(C - 2\ln t)$ , so $y^2 = 1/\bigl(t^2(C - 2\ln t)\bigr)$ wherever $C - 2\ln t > 0$ . The zero function is also a solution and has to be kept separately. Skipping the division by $t$ would put the wrong coefficients into $p$ and $q$ .

Solve $y' = 2y - y^2$ with $y(0) = 1$ .

Show the answer

Written as $y' - 2y = -y^2$ , it is a Bernoulli equation with $n = 2$ . With $z = 1/y$ we get $z' = -y'/y^2 = -2/y + 1$ , so $z' + 2z = 1$ . The integrating factor $e^{2t}$ gives $(e^{2t}z)' = e^{2t}$ , so $z = \tfrac{1}{2} + Ce^{-2t}$ . From $z(0) = 1$ we get $C = \tfrac{1}{2}$ , and

y = \frac{2}{1 + e^{-2t}}.

The denominator is positive for every $t$ , so the solution exists on the whole line, and it rises from $0$ toward the equilibrium $2$ . A direct check works, since $y' = 4e^{-2t}/(1 + e^{-2t})^2$ and $2y - y^2$ simplifies to the same expression.

A classmate solves $y' + y = y^2$ with $z = 1/y$ and reports that every solution has the form $y = 1/(1 + Ce^t)$ and exists for all $t$ . Find the solution with $y(0) = 2$ and its maximal interval, and name the solution that the formula leaves out.

Show the answer

From $z(0) = \tfrac{1}{2} = 1 + C$ we get $C = -\tfrac{1}{2}$ , so

y = \frac{1}{1 - \tfrac{1}{2}e^t} = \frac{2}{2 - e^t}.

The denominator vanishes at $t = \ln 2 \approx 0.693$ , and $y \to +\infty$ as $t$ approaches $\ln 2$ from the left, so the maximal interval is $(-\infty, \ln 2)$ , not the whole line. The formula also leaves out the zero solution $y = 0$ , which satisfies the original equation but corresponds to no value of $C$ , because $z = 1/y$ is undefined there.

Bernoulli Equations

Key terms

Bernoulli equation

Power substitution

Reduction to a linear equation

Exceptional exponents

Zero solution

Check yourself

How this lesson connects

Builds on