Bernoulli’s equation

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 Bernoulli’s equation

The Bernoulli equation is one of the most well-known first-order nonlinear differential equations. It is written as

Bernoulli's equation

where a ( x ) and b ( x ) are continuous functions.

If m = 0, then the Bernoulli equation becomes a linear differential equation. In the case where m = 1, the equation is transformed into an equation with separable variables.

In the general case, when m ≠ 0, 1, the Bernoulli equation is reduced to a linear differential equation using the substitution

change of variable in the Bernoulli equation

The new differential equation for the function z ( x ) is

and can be solved by the methods described on the page Linear Differential Equations of the First Order.

   Example 1

Find the general solution of the equation y’ − y = y^2 e^x.

Solution.

For the given Bernoulli equation, m = 2, so we make the substitution.

      

Differentiating both parts of the equation (the variable y is considered as a complex function of x ), we can write:

      

Let’s divide both parts of the original differential equation by 2 :

      

Substituting z and z’, we find:

      

We have obtained a linear equation for the function z ( x ). Let’s solve it using an integrating factor:

      

The formula expresses the general solution of a linear equation.

      

Returning to the function y ( x ), we obtain the answer in implicit form:

      

which can also be written as:

      

Note that when dividing the equation by 2, we lost the solution y = 0. As a result, the full answer is written as:

      
   Example 2

Solve the differential equation .

Solution.

It is easy to see that this differential equation is a Bernoulli equation. To solve it, we perform the substitution.

      

After differentiation, we obtain:

      

Let’s divide the original equation by 2 and replace y with z :

      

When dividing by 2, we lost the solution y = 0. (This can be verified by direct substitution.)

The differential equation for the new variable z is:

      

We have obtained a linear equation for the function z ( x ), which can be solved, for example, using an integrating factor:

      

It is easy to check that such an integrating factor will be the function 1/ x. Indeed:

      

It is clear that the left side of the equation after multiplying by 1/ x will be the product z ( x ) u ( x )

Then the general solution of the linear differential equation for the function z ( x ) is determined by the formula.

      

Taking into account that y = 1/ z, we write the answer in the form:

      

or implicitly:

      

Therefore, the final answer is:

      
   Example 3

Find all solutions of the differential equation y’ + y cot x = 4 sin x.

Solution.

In this example, we are dealing with the Bernoulli equation with parameter m = 4. Therefore, we make the substitution z = y 1 − m = −3. The derivative will be equal to

      

Let’s multiply both sides of the original equation by (−3) and divide by 4 :

      

Note that when dividing by 4 we lost the solution y = 0. Writing the last equation through the variable z, we get

      

This differential equation is linear. It can be solved, for example, using an integrating factor:

      

We take the function as the integrating factor . After multiplying by u ( x ) , the left side of the equation will be the derivative of the product z ( x ) u ( x ) :

      

Therefore, the general solution of the linear differential equation for the function z ( x ) is represented as:

      

Since z = y −3, we obtain the following solutions to the original Bernoulli equation:

      
   Example 4

Find all solutions of the differential equation .

Solution.

This equation is the Bernoulli equation with a fractional parameter m = 1/2. It can be reduced to a linear differential equation by substitution . The derivative of the new function z ( x ) will be equal to

      

Let’s divide the original Bernoulli equation by . Similar to other examples on this page, the root y = 0 is also a trivial solution to the differential equation. Therefore, we can write:

      

Replacing y with z, we find:

      

So, we have a linear equation for the function z ( x ). The integrating factor here will be equal to

      

Let us choose the function u ( x ) = x as the integrating factor. It can be verified that after multiplying by u ( x ), the left-hand side of the equation will be the derivative of the product z ( x ) u ( x ) :

      

Then the general solution of the linear differential equation will be determined by the expression:

      

Returning to the original function y ( x ), we write the solution in implicit form:

      

So, the full answer looks like this:

      
   Example 5

Find a solution to the differential equation xyy’ = y 2 + 2 that satisfies the initial condition y (1) = 2.

Solution.

First, we check that the given differential equation is a Bernoulli equation:

      

As we can see, we have the Bernoulli equation with parameter m = −1. Therefore, we can make the substitution z = y 1 − m = 2. The derivative will be equal to: z’ = 2 yy’. Next, we multiply both parts of the differential equation by 2 y :

      

Replacing y with z, we transform Bernoulli’s equation into a linear differential equation:

      

Let’s calculate the integrating factor:

      

Let’s find the general solution of the linear equation:

      

Considering that z = y 2, the solution can be written as:

      

Now we define the constant C corresponding to the initial condition y (1) = 2.
Only the solution with a positive sign satisfies this condition. Therefore,

      

As a result, we obtain: C = 4.

Thus, the solution to the Cauchy problem is expressed by the function.

      

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