For an equation of the type \(y' = p(x)y + q(x)y^n\), called a Bernoulli Equation, we can use the special substitution \(v = y^{1n}\), 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.
video by PatrickJMT 

video by Engineer4Free 

video by MIT OCW 

Practice
Unless otherwise instructed, solve these Bernoulli Equations. Give your answers in exact form.
\(y' =(A\cos t+B)yy^3\)
Problem Statement 

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

video by PatrickJMT 

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\(t^2y'+2tyy^3=0\)
Problem Statement 

Solve the Bernoulli Equation \(t^2y'+2tyy^3=0\)
Solution 

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

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 

video by Engineer In Training Exam 

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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You CAN Ace Differential Equations
external links you may find helpful 

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  doubleangle formulas  

\( \sin(2t) = 2\sin(t)\cos(t)\) 
\(\displaystyle{ \cos(2t) = \cos^2(t)  \sin^2(t) }\) 
Set 4  halfangle 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
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\) 
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\)  
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) 
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = \frac{1}{1+t^2} }\)  
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
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.