# Bernoulli Equation

Bernoulli equation is one of the well known nonlinear differential equations of the first order. It is written as

${y’ + a\left( x \right)y }={ b\left( x \right){y^m},}$

where $$a\left( x \right)$$ and $$b\left( x \right)$$ are continuous functions.

If $$m = 0,$$ the equation becomes a linear differential equation. In case of $$m = 1,$$ the equation becomes separable.

In general case, when $$m \ne 0,1,$$ Bernoulli equation can be converted to a linear differential equation using the change of variable

$z = {y^{1 – m}}.$

The new differential equation for the function $$z\left( x \right)$$ has the form:

${z’ + \left( {1 – m} \right)a\left( x \right)z }={ \left( {1 – m} \right)b\left( x \right)}$

and can be solved by the methods described on the page Linear Differential Equation of First Order.

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Find the general solution of the equation $$y’ – y =$$ $${y^2}{e^x}.$$

### Example 2

Solve the differential equation $$y’ + {\large\frac{y}{x}\normalsize}$$ $$= {y^2}.$$

### Example 3

Solve the equation $$y’ + y\cot x$$ $$= {y^4}\sin x.$$

### Example 4

Find all solutions of the differential equation $$y’ + {\large\frac{{2y}}{x}\normalsize}$$ $$= 2x\sqrt y .$$

### Example 5

Find the solution of the differential equation $$4xyy’ = {y^2} + {x^2},$$ satisfying the initial condition $$y\left( 1 \right) = 1.$$

### Example 1.

Find the general solution of the equation $$y’ – y =$$ $${y^2}{e^x}.$$

Solution.

We set $$m = 2$$ for the given Bernoulli equation, so we use the substitution

$z = {y^{1 – m}} = \frac{1}{y}.$

Differentiating both sides of the equation (we consider $$y$$ in the right side as a composite function of $$x$$), we obtain:

$z’ = {\left( {\frac{1}{y}} \right)^\prime } = – \frac{1}{{{y^2}}}y’.$

Divide both sides of the original differential equation by $${y^2}:$$

${y’ – y = {y^2}{e^x},\;\;}\Rightarrow {\frac{{y’}}{{y^2}} – \frac{1}{y} = {e^x}.}$

Substituting $$z$$ and $$z’,$$ we find

$– z – z = {e^x},\;\; \Rightarrow z’ + z = – {e^x}.$

We get the linear equation for the function $$z\left( x \right).$$ To solve it, we use the integrating factor:

$u\left( x \right) = {e^{\int {1dx} }} = {e^x}.$

Then the general solution of the linear equation is given by

${z\left( x \right) }={ \frac{{\int {u\left( x \right)f\left( x \right)dx} + C}}{{u\left( x \right)}} }={ \frac{{\int {{e^x}\left( { – {e^x}} \right)dx} + C}}{{{e^x}}} }={ \frac{{ – \frac{{{e^{2x}}}}{2} + C}}{{{e^x}}} }={ – \frac{{{e^x}}}{2} + C{e^{ – x}} }={ \frac{{2C{e^{ – x}} – {e^x}}}{2}.}$

Since $$C$$ is an arbitrary constant, we can replace $$2C$$ with a constant $$C_1.$$ Returning to the function $$y\left( x \right),$$ we obtain the implicit expression:

${y = \frac{1}{z} }={ \frac{2}{{{C_1}{e^{ – x}} – {e^x}}}.}$

Note that we have lost the solution $$y = 0$$ when dividing the equation by $${y^2}.$$ Thus, the final answer is given by

${y = \frac{2}{{{C_1}{e^{ – x}} – {e^x}}},\;\;}\kern0pt{y = 0.}$

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Problem 1
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Problems 2-5