Bernoulli Equations

Author: John J Weber III, PhD Corresponding Textbook Sections:

• Section 2.6 – Substitutions and Transformations

Expected Educational Results

• Objective 4–1: I can identify the appropriate substitution needed for a given ODE.

• Objective 4–2: I can apply the appropriate substitution to solve an ODE.

Bernoulli Equations

Definition: Bernoulli Equations

A first-order equation that can be written in the form

${\displaystyle \frac{dy}{dx}+P(x)y=Q(x)y^n}$$\displaystyle\frac{dy}{dx}+P(x)y=Q(x)y^n$,

where $P(x)$$P(x)$ and $Q(x)$$Q(x)$ are continuous on an interval $(a,b)$$(a,b)$ and $n$$n$ is a real number, is called a Bernoulli equation.

Method for Solving Bernoulli Equations

1. Use the substitution ${\displaystyle z=y^{1-n}}$$\displaystyle z=y^{1-n}$.

2. Differentiate using chain rule: ${\displaystyle \frac{dz}{dx}=(1-n)y^{-n}\frac{dy}{dx}}$$\displaystyle\frac{dz}{dx}=(1-n)y^{-n}\frac{dy}{dx}$.

3. Multiply both sides of the equation by $(1-n)y^{-n}$$(1-n)y^{-n}$ results in ${\displaystyle (1-n)y^{-n}\frac{dy}{dx}+(1-n)P(x)y^{1-n}=(1-n)Q(x)}$$\displaystyle(1-n)y^{-n}\frac{dy}{dx}+(1-n)P(x)y^{1-n}=(1-n)Q(x)$.

4. Rewrite the DE as ${\displaystyle \frac{dz}{dx}+(1-n)P(x)z=(1-n)Q(x)}$$\displaystyle\frac{dz}{dx}+(1-n)P(x)z=(1-n)Q(x)$ which is first-order linear.

5. Resubstitute $y^{1-n}$$y^{1-n}$ for $z$$z$.

6. Identify missing solutions or solutions that are not solutions of the original DE.

Question 03

Solve the following DEs:

1. ${\displaystyle \frac{dy}{dx}+2xy=e^{3x^2+2x}{y}^4}$$\displaystyle \frac{dy}{dx}+2xy=e^{3x^2+2x}y^4$

2. ${\displaystyle \frac{dy}{dx}=y^4\cos (x)+y\tan (x)}$$\displaystyle \frac{dy}{dx}=y^4\cos{(x)}+y\tan{(x)}$

3. ${\displaystyle x\frac{dy}{dx}+y=\frac{-y^2}{x}}$$\displaystyle x\frac{dy}{dx}+y=\frac{-y^2}{x}$

4. ${\displaystyle x\frac{dy}{dx}-y=e^{x^2}{y}^5}$$\displaystyle x\frac{dy}{dx}-y=e^{x^2}y^5$

5. ${\displaystyle y^{\prime }-\frac{1}{x}y=\frac{1}{2y}}$$\displaystyle y^{\,\prime}-\frac{1}{x}y=\frac{1}{2y}$

6. ${\displaystyle y^{\prime }+\frac{x}{1-x^2}y=x\sqrt{y}}$$\displaystyle y^{\,\prime}+\frac{x}{1-x^2}y=x\sqrt{y}$, $y(0)=1$$y(0)=1$

CC BY-NC-SA 4.0

This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License [http://creativecommons.org/licenses/by-nc-sa/4.0/].

Last Modified: Sunday, 6 September 2020 14:15 EDT