17Calculus Differential Equations - Bernoulli Equation
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.
Recommended Books on Amazon | ||
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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 |
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Engineer4Free - How to solve Bernoulli differential equations [5mins-29secs]
video by Engineer4Free |
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MIT OCW - Bernoulli Equation and example [11mins-36secs]
video by MIT OCW |
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Practice
Unless otherwise instructed, solve these Bernoulli Equations. Give your answers in exact form.
\(y' =(A\cos t+B)y-y^3\)
Problem Statement |
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Solve the Bernoulli Equation \(y' =(A\cos t+B)y-y^3\)
Solution |
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2417 video
video by PatrickJMT |
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\(t^2y'+2ty-y^3=0\)
Problem Statement |
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Solve the Bernoulli Equation \(t^2y'+2ty-y^3=0\)
Solution |
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We include two solutions to this problem by two different instructors.
2418 video
video by PatrickJMT |
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2418 video
video by Engineer4Free |
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\(y'+xy=xy^2\)
Problem Statement |
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Solve the Bernoulli Equation \(y'+xy=xy^2\)
Solution |
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2419 video
video by MIP4U |
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\(\displaystyle{ xy'+y=\frac{1}{y^2} }\)
Problem Statement |
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Solve the Bernoulli Equation \(\displaystyle{ xy'+y=\frac{1}{y^2} }\)
Solution |
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2420 video
video by MIP4U |
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\(\displaystyle{ \frac{dy}{dx} + y = xy^4 }\)
Problem Statement |
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Solve the Bernoulli Equation \(\displaystyle{ \frac{dy}{dx} + y = xy^4 }\)
Solution |
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2421 video
video by Houston Math Prep |
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\(\displaystyle{ \frac{dy}{dx} = \sqrt{y} - y; y(0) = 9 }\)
Problem Statement |
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Solve the Bernoulli Equation \(\displaystyle{ \frac{dy}{dx} = \sqrt{y} - y; y(0) = 9 }\)
Solution |
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2422 video
video by blackpenredpen |
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\(\displaystyle{ \frac{dy}{dx} - \frac{1}{x}y = xy^2 }\)
Problem Statement |
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Solve the Bernoulli Equation \(\displaystyle{ \frac{dy}{dx} - \frac{1}{x}y = xy^2 }\)
Solution |
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2423 video
video by Houston Math Prep |
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\(\displaystyle{ y' + \frac{2}{x}y = -x^9y^5; y(1) = 1 }\)
Problem Statement |
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Solve the Bernoulli Equation \(\displaystyle{ y' + \frac{2}{x}y = -x^9y^5; y(1) = 1 }\)
Solution |
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2424 video
video by MIP4U |
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\( y' = -y(1 + xy^2) \)
Problem Statement |
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Solve the Bernoulli Equation \( y' = -y(1 + xy^2) \)
Solution |
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2425 video
video by blackpenredpen |
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\(\displaystyle{ y' + \frac{4}{x} y = x^3y^2 }\)
Problem Statement |
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Solve the Bernoulli Equation \(\displaystyle{ y' + \frac{4}{x} y = x^3y^2 }\)
Solution |
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This problem was solved by three different people. We include all three videos here.
2426 video
video by The Lazy Engineer |
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2426 video
video by Engineer In Training Exam |
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2426 video
video by MIP4U |
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\( y' + 2xy = xy^3 \)
Problem Statement |
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Solve the Bernoulli Equation \( y' + 2xy = xy^3 \)
Solution |
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2427 video
video by Houston Math Prep |
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\(\displaystyle{ y' + \frac{1}{t}y = -ty^3 }\)
Problem Statement |
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Solve the Bernoulli Equation \(\displaystyle{ y' + \frac{1}{t}y = -ty^3 }\)
Solution |
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2428 video
video by Engineer4Free |
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\(\displaystyle{ \frac{dy}{dt} = 2y + y^5; y(0) = 1 }\)
Problem Statement |
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Solve the Bernoulli Equation \(\displaystyle{ \frac{dy}{dt} = 2y + y^5; y(0) = 1 }\)
Solution |
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2429 video
video by MIP4U |
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You CAN Ace Differential Equations
Topics You Need To Understand For This Page
Related Topics and Links
external links you may find helpful |
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Trig Formulas
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
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Set 1 - basic identities | |||
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\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) |
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) |
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) |
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) |
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Set 2 - squared identities | ||
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\( \sin^2t + \cos^2t = 1\) |
\( 1 + \tan^2t = \sec^2t\) |
\( 1 + \cot^2t = \csc^2t\) |
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Set 3 - double-angle formulas | |
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\( \sin(2t) = 2\sin(t)\cos(t)\) |
\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\) |
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Set 4 - half-angle formulas | |
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\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\) |
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) |
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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Topics Listed Alphabetically
Single Variable Calculus |
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Multi-Variable Calculus |
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Differential Equations |
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Precalculus |
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Practice Instructions
Unless otherwise instructed, solve these Bernoulli Equations. Give your answers in exact form.
