Bernoulli equations
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Topic Notes
Bernoulli Equations
In the past sections we have worked through techniques which solve first order linear differential equations but now the time has come for us to take a look at methodology on how to solve equations that are not linear since realistically speaking, we will not always encounter the same type of equations in our path.
Let us first remind ourselves about the difference between a linear and a nonlinear differential equation. In basic algebra a linear equation is the one in which its independent variable's order is one, in other words, the variable appears, either by itself or accompanied by a constant coefficient where the exponent of the variable is one. And so, a nonlinear equation would be the one where the variable has exponents different than one. For the case of linear differential equations the function can appear by itself, accompanied by coefficients or you can have the derivatives of the function. The equation will be linear as long as the variable and its derivatives have exponents equal to one, if that is not the case, then the equation is nonlinear.
Example:
We may be inclined to rapidly conclude that nonlinear differential equations are much more complicated to solve than linear equations but, do not resent them yet. Although this statement is true for most cases, efficient and intuitive enough mathematical techniques have been developed to resolve certain nonlinear differential equations in a clear and relatively straightforward manner. Thus, you will be pleasantly surprised by the realization that you should already know most of the material that will be covered on this lesson, and that the first technique we would like to study uses a simplification approach, this method makes use of something called Bernoulli equations to transform the original nonlinear equations in a problem into something we can solve right away, and for this reason, the Bernoulli equation method includes just a few more steps than the techniques we have been looking so far to solve linear differential equations.
You will see throughout this article that we will extensively review topics we have seen in past sections since the method to solve a differential equation through the Bernoulli equation is just a mathematical trick to simplify a somewhat unconventional equation into a simpler case we already know how to solve. For that reason, we recommend that if you have not studied the integrating factor method to solve first order exact differential equations yet, you take a look into both sections before you continue with this lesson.
By now, it is also important for you to have mastered skills such as the separable equations technique. If you think you need a quick review on it, do not hesitate to go back and check examples on it.
What is the Bernoulli equation?
A Bernoulli differential equation is a nonlinear first order differential equation which can be reduced to a linear first order differential expression. The special case for these equations is that they have known exact solutions, and after you have done the transformation into the linear expression you can solve the equation by the integrating factor method with exact equations. The general form of a Bernoulli equation looks as follows:
Where n cannot be equal to 0 or 1, since any of these two exponents would result in a linear differential equation expression, and therefore, the Bernoulli equation method would not be needed.
How to solve Bernoulli equations
In order for us to list step by step instructions on how to solve Bernoulli differential equations we will start by using the general form of the equations to give a rough idea of the process, then we will go through a full example that you can also find on the videos for this section.

Starting with equation 2, which is a nonlinear differential expression, we select the highest order term and divide the whole expression by it:

We use a substitution to transform the nonlinear equation into a linear one:

Plugging this into the nonlinear equation will result in a linear differential expression! You can see the development of this substitution below, and notice how the expression has been rearranged to its simplest form for practical purposes:

And we are left with a linear equation of the form
So as you can see the Bernoulli equation method is just a way to transform the nonlinear expression into a linear one, afterwards, we continue to solve the linear differential equation with methodology we have already learned in past sections. There are a few details that need to be taken into account and followed closely while we work through these kind of problems, and therefore, for the purpose of observing the whole process in action, let us work through our first Bernoulli equation example.
Explicitly solve the differential equation:
By finding y as a function of x, having the initial condition: $y(\frac{1}{2})=1$

Step one:
Start by dividing the whole expression by the highest power of y you can find. 
Step two:
In this case n=3, so we set z and z' and substitute them into the nonlinear equation:
Notice this last equation is just the rearranged differential equation after the z substitution. Now you can see that equation for example 1(b) is a linear differential equation of the form: 
Step three:
Having a simpler expression, we can now use any of the two methods we have seen in past sections to find the solution for a linear differential equation. So the question is, how do we pick between the method of exact differential equations, and the integrating factor method?
As a general rule, we suggest you try to solve these problems with the method of exact equations first, if it is not possible to find the solution that way, then move onto using an integrating factor.
So we try to solve equation example 1(b) through the exact equations technique. For that, we need to meet the condition that the partial derivative of M with respect to z, is equal to the partial derivative of N with respect to x.

Step four:
Following the method of integrating factor to find an exact equation to solve for, we multiply equation for example 1(b) for a factor we call "mu of x" (which means this factor denoted by the greek letter "mu" is a function of x).
Now we check once more for the condition $M_z=N_x$ but this time we set it as correct:
In order to continue we need to find out what $\mu$ is.

Step five:
Solve for $\mu$ using the separable differential equations approach:
Applying the exponential to the equation we obtain: $\mu = \frac{1}{x^2}$ 
Step six:
Knowing our integrating factor, we can now multiply it to our equation for example 1(b) and check if we have an exact equation now:

Step seven:
An exact equation is one of the form:
We substitute the expression for z in terms of y: $z=\frac{1}{y^2}$
And we can obtain the final solution to our initial differential equation for example 1:

Step eight:
We input the initial condition $y(\frac{1}{2})=1$ to find the explicit solution of equation (a):
And so, having solved the original nonlinear differential equation for example 1 with the given conditions we find that:
If you got to the end, congratulations! Although a little tedious, you probably noticed that except for the first two steps this wasn't new material and hopefully the methodology made sense. We have summarized all of the steps followed in the last pages for the explicit example in the next list, in that way you have a reference to come and check out quickly while you are doing independent study.
Bernoulli Equation Method, Summarized Steps:
1. Divide the expression given by the highest power of y you can find.
2. Identify n, z and z', and substitute z and z' in the original nonlinear equation to transform it into a linear expression of the form:
3. Solve the linear differential equation by first checking if is an exact equation. The condition for exact equations says that the partial derivative of M with respect to z, is equal to the partial derivative of N with respect to x. If the condition is met, solve the problem through the exact differential equations method.
4. If the condition is not met, multiply the expression by an integrating factor $\mu$
 Identify the new M(x,z) and N(x,z) in the expression and obtain their appropriate partial derivatives and set the condition to make the equation "perfect".
5. Solve for $\mu$ using the separable equations method.
6. Multiply our known expression for $\mu$ and check if it meets the condition of an exact equation.
7. If the condition on top is met, solve the equation through the exact equations method.
 Do not forget to substitute the value of z again in the final solution so you end up with a solution in terms of x and y, instead of x and z.
8. If initial conditions are given, use them to find the explicit solution of the equation.
 Do not forget to solve for y at the end.
Solving Bernoulli differential equations
In this section we will be solving examples of Bernoulli differential equations and how we transform them into linear differential equations. Notice that for each case we will only be going through steps 1 and 2 listed above since those are the steps of the transformation from nonlinear to linear differential equation. The rest of the steps are mostly a review of exact equations, integrating factor and even separable equations techniques, thus we will keep it simple and only work the steps containing the Bernoulli equation method.
For that same reason, the examples do not contain their explicit solutions.
We recommend you practice how to finish the complete solution for these examples by going through the integrating factor method with exact equations until step seven. You cannot go any further since there will be no conditions given for each of these examples, still, the bulk of the process can be done allowing you to restudy through a few past lessons all at once.

Example 2:
Having equation:

Example 3:
Having equation:

Example 4:
Having equation:
We finish this article by recommending you to visit the next differential equations notes on the Bernoulli equation, where you can find information and examples (even graphic representations) to complement your understanding for this lesson.
$\frac{dy}{dx}+P(x)y=Q(x)y^n$
Where both $P(x)$ and $Q(x)$ are both functions of $x$.
To solve this equation first divide the entire equation by $y^n$
$[\frac{dy}{dx}+P(x)y=Q(x) y^n ] \frac{1}{y^n}$
$\frac{1}{y^n} \frac{dy}{dx}+P(x) \frac{1}{y^{n1}} =Q(x)$
And then we can set $z= \frac{1}{y^{n1}} =y^{1n}$
Using implicit differentiation with respect to $x$:
$\frac{1}{(1n)} \frac{dz}{dx}+P(x)z=Q(x)$
or,
$Q(x)P(x)z+ \frac{1}{(n1)} \frac{dz}{dx}=0$
Which we can solve by using our integrating factor technique.
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