# 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

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

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

### Method for Solving Bernoulli Equations

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

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

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

4. Rewrite the DE as $\frac{dz}{dx}+\left(1-n\right)P\left(x\right)z=\left(1-n\right)Q\left(x\right)$$\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. $\frac{dy}{dx}+2xy={e}^{3{x}^{2}+2x}{y}^{4}$$\displaystyle \frac{dy}{dx}+2xy=e^{3x^2+2x}y^4$

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

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

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

5. ${y}^{\phantom{\rule{0.167em}{0ex}}\prime }-\frac{1}{x}y=\frac{1}{2y}$$\displaystyle y^{\,\prime}-\frac{1}{x}y=\frac{1}{2y}$

6. ${y}^{\phantom{\rule{0.167em}{0ex}}\prime }+\frac{x}{1-{x}^{2}}y=x\sqrt{y}$$\displaystyle y^{\,\prime}+\frac{x}{1-x^2}y=x\sqrt{y}$, $y\left(0\right)=1$$y(0)=1$