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 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{{ – 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

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

    Page 1
    Problem 1
    Page 2
    Problems 2-5