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 (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 \[y' - y = {y^2}{e^x}.\]


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 \[y' + \frac{y}{x} = {y^2}.\]


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.

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