Bernoulli Differential Equations for Beginners: Everything You Need to Know
This guide on Bernoulli differential equations for beginners will help you understand one of the most important classes of first-order nonlinear differential equations. While they may initially appear difficult due to their nonlinear form, they can be transformed into linear differential equations with a simple substitution, making them much simpler to solve. Because of this feature, Bernoulli differential equations are often among the nonlinear differential equations studied in differential equations.
In this guide, you will learn what Bernoulli differential equations are and how to identify them. We will derive the solution method step by step, explain the substitution that transforms the equation into a linear form, and provide a systematic procedure for solving these equations. Finally, we will work through several examples with detailed solutions.
By the end of this article, you will have everything you need to understand and solve Bernoulli differential equations.
What Are Bernoulli Differential Equations?
A Bernoulli differential equation is a first-order differential equation that can be written in the form
$$\frac{dy}{dx} + p(x)y = g(x)y^n,$$
where \( p(x) \) and \( g(x) \) are functions of \( x \), and \( n \) is a real number satisfying \( n \neq 0 \) and \( n \neq 1 \).
These equations are important because they provide one of the few examples of nonlinear differential equations that can still be solved analytically.
Deriving the Solution Method for Bernoulli Differential Equations
Bernoulli equations have a special structure that allows us to transform them into linear differential equations through a change of variables. This idea should feel familiar if you have read the articles The Ultimate Step-by-Step Guide to Solving Integrals Using Substitution or Beginner’s Guide to Changing Variables in Double and Triple Integrals: Everything You Need to Know. In each case, the goal is the same: replace a difficult expression with a new variable that simplifies the problem.
Let’s begin with the standard form of a Bernoulli differential equation:
$$\frac{dy}{dx} + p(x)y = g(x)y^n,$$
where \( n \neq 0 \) and \( n \neq 1 \). Dividing every term by \( y^n \) gives
$$y^{-n}\frac{dy}{dx} + p(x)y^{1-n} = g(x).$$
Next, we let
$$v = y^{1-n}.$$
Next, we need to differentiate v with respect to x. For a review of differentiation, please refer to the article How to Differentiate a Function Step by Step: A Beginner’s Guide. Applying the chain rule, we find
$$\frac{dv}{dx} = (1 – n)y^{-n}\frac{dy}{dx}.$$
Solving for \( y^{-n}\frac{dy}{dx} \) gives
$$\frac{1}{1 – n}\frac{dv}{dx} = y^{-n}\frac{dy}{dx}.$$
Substituting into the equation, we have
$$\frac{1}{1 – n}\frac{dv}{dx} + p(x)v = g(x).$$
To simplify the equation, multiply both sides by \( 1 – n \) to get
$$\frac{dv}{dx} + (1 – n)p(x)v = (1-n)g(x).$$
The transformed equation is now a linear differential equation in standard form. It can be solved using the integrating factor method developed in A Step-by-Step Tutorial on Linear Differential Equations with Examples and Solutions.
Once the linear differential equation is solved, don’t forget to back-substitute.
Now that we have derived the solution method, we can summarize the procedure into a e sequence of steps that can be applied to any Bernoulli differential equation.
Steps to Solve Bernoulli Differential Equations
Now that we have derived the solution method, let’s summarize the process into a series of steps that can be applied to any Bernoulli differential equation.
Step 1: Write the Equation in Standard Form
Begin by rewriting the differential equation in the standard form
$$\frac{dy}{dx} + p(x)y = g(x)y^n.$$
Step 2: Divide by \( y^n \)
Divide every term in the equation by \( y^n\ \).
Step 3: Prepare the Substitution
Introduce the substitution
$$v = y^{1-n}.$$
Next, differentiate both sides:
$$\frac{dv}{dx} = (1-n)y^{-n}\frac{dy}{dx}.$$
Solving for \( y^{-n}\frac{dy}{dx} \) gives
$$y^{-n}\frac{dy}{dx} = \frac{1}{1-n}\frac{dv}{dx}.$$
Step 4: Make the Substitution
Substitute
$$v = y^{1-n}$$
and
$$\frac{1}{1 – n}\frac{dv}{dx} = y^{-n}\frac{dy}{dx}$$
into the equation.
After simplifying, you will obtain
$$\frac{dv}{dx} + (1-n)p(x)v = (1 – n)g(x).$$
Step 5: Solve the Resulting Linear Differential Equation
Since the equation is now linear, the integrating factor method can be used.
Step 6: Convert Back to the Original Variable
Back substitute and solve for y.
This solves the original Bernoulli differential equation.
Step 7: Apply the Initial Condition (If Given)
If an initial condition is provided, substitute it into the general solution to determine the constant \( C \).
Worked Out Examples
Now that we understand the theory behind Bernoulli differential equations and have developed a step-by-step procedure for solving them, let’s work through a couple of examples. The examples require the techniques covered in the articles Basic Integration Problems for Beginners and Integration by Parts Explained with Examples: A Step-by-Step Guide.
Example 1: Solve \( \frac{dy}{dx} – y = xy^2 \).
Solution: The equation is already in standard form. Multiplying each term of the equation by \( y^{-2} \), we obtain
$$y^{-2}\frac{dy}{dx} – y^{-1} = x.$$
Next we let
$$v = y^{-1}.$$
Using the chain rule to differentiate this, we get
$$\frac{dv}{dx} = -y^{-2}\frac{dy}{dx}.$$
Multiplying both sides by -1 then gives
$$-\frac{dv}{dx} = y^{-2}\frac{dy}{dx}.$$
Substituting into the differential equation, we find
$$-\frac{dv}{dx} – v = x.$$
Multiplying each term of the equation by \( -1 \), we obtain
$$\frac{dv}{dx} + v = x.$$
The integrating factor is
$$\mu(x) = e^{\int dx}.$$
Integrating we get
$$\mu(x) = e^x.$$
Multiplying each term of the equation by the integrating factor, we obtain
$$e^x\frac{dv}{dx} + e^xv = xe^x.$$
Rewriting the left-hand side as a product rule, this becomes
$$\frac{d}{dx}(e^xv) = xe^x.$$
Integrating both sides, we find
$$\int \frac{d}{dx}(e^xv) dx = \int xe^x dx.$$
To evaluate the integral on the right-hand side, choose \( u = x \) and \( dv = e^xdx \), then \( du = dx \) and \( v = e^x \). Applying the integration by parts formula then gives
$$\int \frac{d}{dx}(e^xv) dx = xe^x – \int e^x dx.$$
Therefore,
$$e^xv = xe^x – e^x + C.$$
Multiplying each term of the equation by \( e^{-x} \) we obtain
$$e^xe^{-x}v = xe^xe^{-x} – e^xe^{-x} + Ce^{-x}.$$
Using exponential properties, we find
$$e^{x – x}v = xe^{x – x} – e^{x – x} + Ce^{-x}.$$
Simplifying this is equivalent to
$$e^0v = xe^0 – e^0 + Ce^{-x}.$$
Hence,
$$1v = 1x – 1 + Ce^{-x}.$$
Which implies
$$v = x – 1 + Ce^{-x}.$$
Back substituting gives
$$y^{-1} = x – 1 + Ce^{-x}.$$
Using exponential properties, this is
$$\frac{1}{y} = x – 1 + Ce^{-x}.$$
Our final solution is then
$$y = \frac{1}{x – 1 + Ce^{-x}}.$$
The next example shows that the method also works with fractional exponents.
Example 2: Solve \( \frac{dy}{dx} + 2y = 4y^{\frac{1}{2}} \).
Solution: The equation is already in standard form. Multiplying each term of the equation by \( y^{-\frac{1}{2}} \), we obtain
$$y^{-\frac{1}{2}}\frac{dy}{dx} + 2y^{\frac{1}{2}} = 4.$$
Next we let
$$v = y^{\frac{1}{2}}.$$
Using the chain rule to differentiate this, we get
$$\frac{dv}{dx} = \frac{1}{2}y^{-\frac{1}{2}}\frac{dy}{dx}.$$
Multiplying both sides by 2 then gives
$$2\frac{dv}{dx} = y^{-\frac{1}{2}}\frac{dy}{dx}.$$
Substituting into the differential equation, we find
$$2\frac{dv}{dx} + 2v = 4.$$
Dividing each term of the equation by \( 2 \), we obtain
$$\frac{dv}{dx} + v = 2.$$
The integrating factor is
$$\mu(x) = e^{\int dx}.$$
Integrating we get
$$\mu(x) = e^x.$$
Multiplying each term of the equation by the integrating factor, we obtain
$$e^x\frac{dv}{dx} + e^xv = 2e^x.$$
Rewriting the left-hand side as a product rule, this becomes
$$\frac{d}{dx}(e^xv) = 2e^x.$$
Integrating both sides, we find
$$\int \frac{d}{dx}(e^xv) dx = \int 2e^x dx.$$
Using the constant multiple rule, this is
$$\int \frac{d}{dx}(e^xv) dx = 2\int e^x dx.$$
Therefore,
$$e^xv = 2e^x + C.$$
Multiplying each term of the equation by \( e^{-x} \) we obtain
$$e^xe^{-x}v = 2e^xe^{-x} + Ce^{-x}.$$
Using exponential properties, we find
$$e^{x – x}v = 2e^{x – x} + Ce^{-x}.$$
Simplifying this is equivalent to
$$e^0v = 2e^0 + Ce^{-x}.$$
Hence,
$$1v = 2(1) + Ce^{-x}.$$
Which implies
$$v = 2 + Ce^{-x}.$$
Back substituting gives
$$y^{\frac{1}{2}} = 2 + Ce^{-x}.$$
Our final solution is then
$$y = (2 + Ce^{-x})^2.$$
Conclusion
In this guide, we explored Bernoulli differential equations for beginners and how they can be transformed into linear differential equations through a carefully chosen substitution. We derived the solution method from first principles, developed a step-by-step procedure for solving these equations, and worked through a couple of examples to demonstrate the method in action.
Further Reading
Separable Differential Equations Explained Step by Step: A Complete Beginner’s Guide – Many differential equations solved using substitution reduce to separable equations rather than linear equations.
