Differential Equations

First Order Equations

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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 \(y’ – y =\) \({y^2}{e^x}.\)


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