Differential Equations

First Order Equations

1st Order Diff Equations Logo

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:

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

Thus, the final answer is

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

See more problems on Page 2.

Page 1 Page 2