## 17Calculus Differential Equations - Bernoulli Equation

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For an equation of the type $$y' = p(x)y + q(x)y^n$$, called a Bernoulli Equation, we can use the special substitution $$v = y^{1-n}$$, which will turn the equation into a linear equation.
Note: This technique uses integrating factors in order to solve the resulting linear equation.

Here are several good videos explaining the theory of how and why this substitution works.

### PatrickJMT - Bernoulli Equation for Differential Equations , Part 1 [10mins-25secs]

video by PatrickJMT

### Engineer4Free - How to solve Bernoulli differential equations [5mins-29secs]

video by Engineer4Free

### MIT OCW - Bernoulli Equation and example [11mins-36secs]

video by MIT OCW

Practice

Unless otherwise instructed, solve these Bernoulli Equations. Give your answers in exact form.

$$y' =(A\cos t+B)y-y^3$$

Problem Statement

Solve the Bernoulli Equation $$y' =(A\cos t+B)y-y^3$$

Solution

### 2417 video

video by PatrickJMT

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$$t^2y'+2ty-y^3=0$$

Problem Statement

Solve the Bernoulli Equation $$t^2y'+2ty-y^3=0$$

Solution

We include two solutions to this problem by two different instructors.

### 2418 video

video by PatrickJMT

### 2418 video

video by Engineer4Free

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$$y'+xy=xy^2$$

Problem Statement

Solve the Bernoulli Equation $$y'+xy=xy^2$$

Solution

### 2419 video

video by MIP4U

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$$\displaystyle{ xy'+y=\frac{1}{y^2} }$$

Problem Statement

Solve the Bernoulli Equation $$\displaystyle{ xy'+y=\frac{1}{y^2} }$$

Solution

### 2420 video

video by MIP4U

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$$\displaystyle{ \frac{dy}{dx} + y = xy^4 }$$

Problem Statement

Solve the Bernoulli Equation $$\displaystyle{ \frac{dy}{dx} + y = xy^4 }$$

Solution

### 2421 video

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$$\displaystyle{ \frac{dy}{dx} = \sqrt{y} - y; y(0) = 9 }$$

Problem Statement

Solve the Bernoulli Equation $$\displaystyle{ \frac{dy}{dx} = \sqrt{y} - y; y(0) = 9 }$$

Solution

### 2422 video

video by blackpenredpen

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$$\displaystyle{ \frac{dy}{dx} - \frac{1}{x}y = xy^2 }$$

Problem Statement

Solve the Bernoulli Equation $$\displaystyle{ \frac{dy}{dx} - \frac{1}{x}y = xy^2 }$$

Solution

### 2423 video

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$$\displaystyle{ y' + \frac{2}{x}y = -x^9y^5; y(1) = 1 }$$

Problem Statement

Solve the Bernoulli Equation $$\displaystyle{ y' + \frac{2}{x}y = -x^9y^5; y(1) = 1 }$$

Solution

### 2424 video

video by MIP4U

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$$y' = -y(1 + xy^2)$$

Problem Statement

Solve the Bernoulli Equation $$y' = -y(1 + xy^2)$$

Solution

### 2425 video

video by blackpenredpen

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$$\displaystyle{ y' + \frac{4}{x} y = x^3y^2 }$$

Problem Statement

Solve the Bernoulli Equation $$\displaystyle{ y' + \frac{4}{x} y = x^3y^2 }$$

Solution

This problem was solved by three different people. We include all three videos here.

### 2426 video

video by MIP4U

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$$y' + 2xy = xy^3$$

Problem Statement

Solve the Bernoulli Equation $$y' + 2xy = xy^3$$

Solution

### 2427 video

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$$\displaystyle{ y' + \frac{1}{t}y = -ty^3 }$$

Problem Statement

Solve the Bernoulli Equation $$\displaystyle{ y' + \frac{1}{t}y = -ty^3 }$$

Solution

### 2428 video

video by Engineer4Free

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$$\displaystyle{ \frac{dy}{dt} = 2y + y^5; y(0) = 1 }$$

Problem Statement

Solve the Bernoulli Equation $$\displaystyle{ \frac{dy}{dt} = 2y + y^5; y(0) = 1 }$$

Solution

### 2429 video

video by MIP4U

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You CAN Ace Differential Equations

 integrating factors - linear

### Trig Formulas

The Unit Circle

The Unit Circle [wikipedia] Basic Trig Identities

Set 1 - basic identities

$$\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }$$

$$\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }$$

$$\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }$$

$$\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }$$

Set 2 - squared identities

$$\sin^2t + \cos^2t = 1$$

$$1 + \tan^2t = \sec^2t$$

$$1 + \cot^2t = \csc^2t$$

Set 3 - double-angle formulas

$$\sin(2t) = 2\sin(t)\cos(t)$$

$$\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }$$

Set 4 - half-angle formulas

$$\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }$$

$$\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }$$

Trig Derivatives

 $$\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }$$ $$\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }$$ $$\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }$$ $$\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }$$ $$\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }$$ $$\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }$$

Inverse Trig Derivatives

 $$\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }$$ $$\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }$$

Trig Integrals

 $$\int{\sin(x)~dx} = -\cos(x)+C$$ $$\int{\cos(x)~dx} = \sin(x)+C$$ $$\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C$$ $$\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C$$ $$\int{\sec(x)~dx} =$$ $$\ln\abs{\sec(x)+\tan(x)}+C$$ $$\int{\csc(x)~dx} =$$ $$-\ln\abs{\csc(x)+\cot(x)}+C$$

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