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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.

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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.

Practice 2417
Solution

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

Problem Statement

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

Solution

2417 video

video by PatrickJMT

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Practice 2418
Solution

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

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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Practice 2419
Solution

$y'+xy=xy^2$

Problem Statement

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

Solution

2419 video

video by MIP4U

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Practice 2420
Solution

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

Problem Statement

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

Solution

2420 video

video by MIP4U

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Practice 2421
Solution

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

Problem Statement

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

Solution

2421 video

video by Houston Math Prep

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Practice 2422
Solution

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

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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Practice 2423
Solution

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

Problem Statement

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

Solution

2423 video

video by Houston Math Prep

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Practice 2424
Solution

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

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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Practice 2425
Solution

$ 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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Practice 2426
Solution

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

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 The Lazy Engineer

2426 video

video by Engineer In Training Exam

2426 video

video by MIP4U

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Practice 2427
Solution

$ y' + 2xy = xy^3 $

Problem Statement

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

Solution

2427 video

video by Houston Math Prep

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Practice 2428
Solution

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

Problem Statement

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

Solution

2428 video

video by Engineer4Free

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Practice 2429
Solution

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

Problem Statement

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

Solution

2429 video

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You CAN Ace Differential Equations

Topics You Need To Understand For This Page

basics of differential equations

substitution

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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Practice Instructions

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

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

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

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