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 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{{ – x + C}}{{{e^x}}} }

= {\left( {C – x} \right){e^{ – x}}.}

\]

Returning to the function \(y\left( x \right),\) we obtain the implicit expression:

\[y = \frac{1}{z} = \frac{1}{{\left( {C – x} \right){e^{ – x}}}},\]

which can be written in the form:

\[y\left( {C – x} \right) = {e^x}.\]

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

\[{y\left( {C – x} \right) = {e^x},\;\;}\kern-0.3pt{y = 0.}\]