# Differential Equations

## First Order Equations # 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 (x) and b (x) 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

### 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

$- z^\prime - z = {e^x},\;\; \Rightarrow z' + z = - {e^x}.$

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

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

### Example 2.

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

Solution.

As it can be seen, this differential equation is a Bernoulli equation. To solve it, we make the substitution

$z = {y^{1 - m}} = \frac{1}{y}.$

Differentiating, we find:

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

Divide the original equation by $${y^2}$$ and replace $$y$$ with $$z:$$

$\frac{{y'}}{{{y^2}}} + \frac{1}{{yx}} = 1.$

When dividing by $${y^2},$$ we have lost the solution $$y = 0.$$ (You can check this by direct substitution.)

In terms of $$z,$$ the differential equation is written in the form:

$- z' + \frac{z}{x} = 1\;\;\text{or}\;\; z' - \frac{z}{x} = - 1.$

We get the linear equation for the function $$z\left( x \right),$$ so we can solve it using the integrating factor technique:

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

We can make sure that the function $$\frac{1}{x}$$ is the integrating factor. Indeed:

$z' \cdot \frac{1}{x} - \frac{z}{x} \cdot \frac{1}{x} = z' \cdot \frac{1}{x} - \frac{z}{{{x^2}}} = {\left( {z \cdot \frac{1}{x}} \right)^\prime }.$

We see that the left side of the equation becomes the derivative of the product $$z\left( x \right)u\left( x \right)$$ after multiplying by $$\frac{1}{x}$$.

Then the general solution of the linear equation for $$z\left( x \right)$$ is given by

$z = \frac{{\int {u\left( x \right)f\left( x \right)dx} + C}}{{u\left( x \right)}} = \frac{{\int {\frac{1}{x} \cdot \left( { - 1} \right)dx} + C}}{{\frac{1}{x}}} = \frac{{ - \ln \left| x \right| + C}}{{\frac{1}{x}}} = x\left( {C - \ln \left| x \right|} \right).$

Taking into account that $$y = \frac{1}{z},$$ we can write the answer:

$y = \frac{1}{{x\left( {C - \ln \left| x \right|} \right)}},$

or in the implicit form:

$yx\left( {C - \ln \left| x \right|} \right) = 1.$

$yx\left( {C - \ln \left| x \right|} \right) = 1,\;\; y = 0.$