Solution 1. order differential equation

santais
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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:

<br /> <br /> y&#039;(t)+p*y(t)=y(t)^2<br /> <br />

Homework Equations



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

y&#039;(x) +p(x)y = q(x)

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:

e^{-µ(x)} * ∫e^{µ(x)} q(x)dx

where µ(x) = ∫p(x)dx

and µ(x) = px
 
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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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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