Differential Equations: Bernoulli Equation

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SUMMARY

The discussion centers on solving the Bernoulli differential equation represented as y' - 3y = y². The user attempts to transform the equation by dividing both sides by y² and substituting v = y^(-1). The correct transformation leads to the first-order ordinary differential equation (ODE) v' + 3v = -1. The final solution is derived as 1/((Ce^(-3x)) - 1/3), confirming that the solution is indeed the reciprocal of the original variable y.

PREREQUISITES
  • Understanding of Bernoulli differential equations
  • Knowledge of first-order ordinary differential equations (ODEs)
  • Familiarity with substitution methods in differential equations
  • Basic calculus, specifically differentiation and integration techniques
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  • Study the method of solving Bernoulli equations in detail
  • Learn about the application of substitution techniques in differential equations
  • Explore the concept of integrating factors for first-order ODEs
  • Review examples of solving differential equations with variable separable methods
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Students studying differential equations, mathematics educators, and anyone looking to deepen their understanding of Bernoulli equations and their solutions.

dmoney123
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Homework Statement


Find the general solution:

y'-3y=(y^2)

Homework Equations

The Attempt at a Solution



divide both sides by y^2

y'(y^-2) -3(y^-1) = 1

we know v=y^(n-1)
v=y^-1
v'=d/dx(y^-1)
v'=-(y^-2) y'

plug it back into

y'(y^-2) -3(y^-1) = 1

-v'-3v=1

this is where I think I am making a mistake

im putting it into a 1st ODE by making v' positive

v'+3v=-1

then u(x)=e^(3x)

=-{[e^(3x)]/3+C}/e^(3x)

=-1/3+C/e^(3x)

the solution given to me is

1/((Ce^(-3x))-1/3)

Any help is appreciated. thanks!
 
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dmoney123 said:

Homework Statement


Find the general solution:

y'-3y=(y^2)

Homework Equations

The Attempt at a Solution



divide both sides by y^2

y'(y^-2) -3(y^-1) = 1

we know v=y^(n-1)
v=y^-1
v'=d/dx(y^-1)
v'=-(y^-2) y'

plug it back into

y'(y^-2) -3(y^-1) = 1

-v'-3v=1

this is where I think I am making a mistake

im putting it into a 1st ODE by making v' positive

v'+3v=-1

then u(x)=e^(3x)

=-{[e^(3x)]/3+C}/e^(3x)

=-1/3+C/e^(3x)

which is equal to ##-\frac 1 3 + Ce^{-3x}##

the solution given to me is

1/((Ce^(-3x))-1/3)

Any help is appreciated. thanks!

Isn't that the reciprocal of what you have? Remember ##v = \frac 1 y##.
 
i always feel so stupid when i come on the site. yes it is the reciprocal. thank you for your help
 

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