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.
video by PatrickJMT |
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video by Engineer4Free |
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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
Solve the Bernoulli Equation \(y' =(A\cos t+B)y-y^3\)
Solution
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.
video by PatrickJMT |
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video by Engineer4Free |
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\(y'+xy=xy^2\)
Problem Statement
Solve the Bernoulli Equation \(y'+xy=xy^2\)
Solution
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
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
video by Houston Math Prep |
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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
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
video by Houston Math Prep |
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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
video by MIP4U |
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\( y' = -y(1 + xy^2) \)
Problem Statement
Solve the Bernoulli Equation \( y' = -y(1 + xy^2) \)
Solution
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.
video by The Lazy Engineer |
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video by Engineer In Training Exam |
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video by MIP4U |
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\( y' + 2xy = xy^3 \)
Problem Statement
Solve the Bernoulli Equation \( y' + 2xy = xy^3 \)
Solution
video by Houston Math Prep |
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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
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
video by MIP4U |
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Really UNDERSTAND Differential Equations
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