Solution 1. order differential equation

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SUMMARY

The discussion centers on solving the first-order differential equation y'(t) + p*y(t) = y(t)^2. The user struggles to rewrite this equation in the standard form of a first-order differential equation, specifically y'(x) + p(x)y = q(x). The confusion arises from the presence of y(t) squared, which complicates finding a general solution using the method of integrating factors. The user references the Bernoulli differential equation and attempts to apply the general solution formula, but encounters difficulties due to the nature of the equation.

PREREQUISITES
  • Understanding of first-order differential equations
  • Familiarity with Bernoulli differential equations
  • Knowledge of integrating factors
  • Basic calculus concepts, including integration
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  • Study the method for solving Bernoulli differential equations
  • Learn about integrating factors in detail
  • Explore specific examples of first-order differential equations
  • Investigate the implications of non-linear differential equations
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Students studying differential equations, educators teaching calculus, and mathematicians interested in non-linear dynamics.

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Homework Statement


So I've been given an assignment to find all solutions to the differential equation as mentioned below. From what can be seen, it's a 1. order differerential equation.

The assignment is as stated:

[itex] <br /> y'(t)+p*y(t)=y(t)^2<br /> [/itex]

Homework Equations



So I tried to rewrite to somehow match the general form of a 1. order differential equation:

[itex]y'(x) +p(x)y = q(x)[/itex]

But no matter what I try, I can't get it to look somehow like it.

The Attempt at a Solution



The problem is that it equals the funktion itself raised in 2. I just have no idea how to find the solution, when that is the case. I tried to rewrite and solve it, using the general solution, but no matter what, the function itself becomes a part of the solution, which shouldn't be the case.

Been using the general solution as mentioned below:

[itex]e^{-µ(x)} * ∫e^{µ(x)} q(x)dx[/itex]

where [itex]µ(x) = ∫p(x)dx[/itex]

and [itex]µ(x) = px[/itex]
 
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"e^{-µ(x)} * ∫e^{µ(x)} q(x)dx,µ(x) = ∫p(x)dx,µ(x) = px" You said µ(x)=px, but p is a function of x, so I believe it's something else. Also I think you meant not the first order since the original diff. equ is already of first order but of first degree.
 

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