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September 2nd, 2018, 01:14 PM   #11
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Quote:
Originally Posted by Integrator View Post
I think that $\displaystyle μ_{0} = e^{x + c}$.
An idea:
It is observed that $\displaystyle f (x) = g (e^{x+c})$...

All the best,

Integrator
c will be cancelled if you multiply both sides of the equation with μ0.
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September 2nd, 2018, 02:04 PM   #12
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So back to the problem :

$\displaystyle
f(x) - \frac{df(x)}{dx} + f^{2}(x) = 0
\Leftrightarrow
-\frac{df(x)}{dx} + f(x) = -f^{2}(x)
$

$\displaystyle
let:f(x) = u^{\frac{1}{1-2}}(x) = u^{-1}(x)
$

$\displaystyle
-\frac{du^{-1}(x)}{dx} + u^{-1}(x) = -u^{-2}(x)
\Leftrightarrow
u^{-2}(x) \cdot \frac{du(x)}{dx} + u^{-1}(x) = -u^{-2}(x)
\Leftrightarrow
\frac{u^{-2}(x)}{u^{-2}(x)} \cdot \frac{du(x)}{dx} + \frac{u^{-1}(x)}{u^{-2}(x)} = \frac{-u^{-2}(x)}{u^{-2}(x)}
\Leftrightarrow
\frac{du(x)}{dx} + u(x) = -1
$

$\displaystyle
let:μ_{0} = e^{\int 1dx} = e^{x+c}
$

$\displaystyle
\frac{du(x)}{dx} \cdot e^{x} + u(x) \cdot e^{x} = -e^{x}
\Leftrightarrow
\frac{d[u(x) \cdot e^{x}]}{dx} = -e^{x}
\Leftrightarrow
\int \frac{d[u(x) \cdot e^{x}]}{dx} \cdot dx = \int -e^{x} dx
\Leftrightarrow
u(x) \cdot e^{x} + c_{1} = \int -e^{x} dx
\Leftrightarrow
u(x) \cdot e^{x} + c_{1} = -e^{x} - c_{2}
\Leftrightarrow
u(x) = \frac{-e^{x}}{e^{x}} - \frac{c_{2} + c_{1}}{e^{x}}
\Leftrightarrow
u(x) = -1 - ke^{-x} : where : k = (c_{2} + c_{1})
$

$\displaystyle
f(x) = u^{-1}(x)
\Leftrightarrow
f(x) = (-1 - ke^{-x})^{-1}
$

Finally:
$\displaystyle
f(x) = \frac{1}{-1 - ke^{-x}}
$
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September 2nd, 2018, 09:28 PM   #13
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"babaliaris",

Very nice and fair! I hope you know what is the second family of functions $\displaystyle f(x)$ resulting from the explanation of the module ...

All the best,

Integrator
Thanks from babaliaris

Last edited by skipjack; September 3rd, 2018 at 12:33 AM.
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September 3rd, 2018, 12:32 AM   #14
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There's no need for substitutions.
If f(x) isn't zero, multiplying by $\displaystyle \frac{e^x}{f^2(x)}$ gives $\displaystyle -\frac{e^x}{f^2(x)}\cdot\frac{df(x)}{dx} + \frac{e^x}{f(x)} = -e^x$,
integrating which gives $\displaystyle \frac{e^x}{f(x)} = -e^x + k$, where $k$ is a constant,
so (if $x = \ln k$ is excluded) $\displaystyle f(x) = \frac{e^x}{-e^x + k} = \frac{1}{-1 + ke^{-x}}$.

However, this function doesn't satisfy the equation given in the original post.
Thanks from babaliaris
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September 3rd, 2018, 05:44 AM   #15
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Quote:
Originally Posted by skipjack View Post
There's no need for substitutions.
If f(x) isn't zero, multiplying by $\displaystyle \frac{e^x}{f^2(x)}$ gives $\displaystyle -\frac{e^x}{f^2(x)}\cdot\frac{df(x)}{dx} + \frac{e^x}{f(x)} = -e^x$,
integrating which gives $\displaystyle \frac{e^x}{f(x)} = -e^x + k$, where $k$ is a constant,
so (if $x = \ln k$ is excluded) $\displaystyle f(x) = \frac{e^x}{-e^x + k} = \frac{1}{-1 + ke^{-x}}$.

However, this function doesn't satisfy the equation given in the original post.
Well this shows you have more experience.

Yes it doesn't, I just split it to conditions, but you still need to get rid of that absolute value - you need more information I guess.

Last edited by skipjack; September 7th, 2018 at 04:47 AM.
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September 3rd, 2018, 06:05 AM   #16
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Another thing I wanted to mention is, what if you square both sides of the original equation? Will this help? I believe this will help with the absolute value, but then the equation becomes extremely difficult to solve.

$\displaystyle
[|f(x)-f'(x)|+(f(x))^2]^{2}=0
$.

Last edited by skipjack; September 7th, 2018 at 04:47 AM.
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September 3rd, 2018, 06:16 AM   #17
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As already posted, the original equation's solution is f(x) = 0.
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September 3rd, 2018, 06:30 AM   #18
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Quote:
Originally Posted by skipjack View Post
There's no need for substitutions.
If f(x) isn't zero, multiplying by $\displaystyle \frac{e^x}{f^2(x)}$ gives $\displaystyle -\frac{e^x}{f^2(x)}\cdot\frac{df(x)}{dx} + \frac{e^x}{f(x)} = -e^x$,
integrating which gives $\displaystyle \frac{e^x}{f(x)} = -e^x + k$, where $k$ is a constant,
so (if $x = \ln k$ is excluded) $\displaystyle f(x) = \frac{e^x}{-e^x + k} = \frac{1}{-1 + ke^{-x}}$.

However, this function doesn't satisfy the equation given in the original post.
Hello,

I think that the family of functions found by "babaliaris" must only verify the differential equation $\displaystyle f(x)-f'(x)+(f(x))^2=0$ because it put the condition $\displaystyle f(x)-f'(x)>0$.

All the best,

Integrator
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September 3rd, 2018, 06:42 AM   #19
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Quote:
Originally Posted by babaliaris View Post
So back to the problem :

$\displaystyle
f(x) - \frac{df(x)}{dx} + f^{2}(x) = 0
\Leftrightarrow
-\frac{df(x)}{dx} + f(x) = -f^{2}(x)
$

$\displaystyle
let:f(x) = u^{\frac{1}{1-2}}(x) = u^{-1}(x)
$

$\displaystyle
-\frac{du^{-1}(x)}{dx} + u^{-1}(x) = -u^{-2}(x)
\Leftrightarrow
u^{-2}(x) \cdot \frac{du(x)}{dx} + u^{-1}(x) = -u^{-2}(x)
\Leftrightarrow
\frac{u^{-2}(x)}{u^{-2}(x)} \cdot \frac{du(x)}{dx} + \frac{u^{-1}(x)}{u^{-2}(x)} = \frac{-u^{-2}(x)}{u^{-2}(x)}
\Leftrightarrow
\frac{du(x)}{dx} + u(x) = -1
$

$\displaystyle
let:μ_{0} = e^{\int 1dx} = e^{x+c}
$

$\displaystyle
\frac{du(x)}{dx} \cdot e^{x} + u(x) \cdot e^{x} = -e^{x}
\Leftrightarrow
\frac{d[u(x) \cdot e^{x}]}{dx} = -e^{x}
\Leftrightarrow
\int \frac{d[u(x) \cdot e^{x}]}{dx} \cdot dx = \int -e^{x} dx
\Leftrightarrow
u(x) \cdot e^{x} + c_{1} = \int -e^{x} dx
\Leftrightarrow
u(x) \cdot e^{x} + c_{1} = -e^{x} - c_{2}
\Leftrightarrow
u(x) = \frac{-e^{x}}{e^{x}} - \frac{c_{2} + c_{1}}{e^{x}}
\Leftrightarrow
u(x) = -1 - ke^{-x} : where : k = (c_{2} + c_{1})
$

$\displaystyle
f(x) = u^{-1}(x)
\Leftrightarrow
f(x) = (-1 - ke^{-x})^{-1}
$

Finally:
$\displaystyle
f(x) = \frac{1}{-1 - ke^{-x}}
$
Hello,

In this case, to complete the differential equation solving we must answer the question:
For which values of the constant $\displaystyle k$ and for which values of $\displaystyle x$ the family of functions $\displaystyle f(x) = \frac{1}{-1 - ke^{-x}}$ is defined in the set $\displaystyle \mathbb R$?

All the best,

Integrator

Last edited by Integrator; September 3rd, 2018 at 06:50 AM.
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September 3rd, 2018, 07:01 AM   #20
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Quote:
Originally Posted by skipjack View Post
As already posted, the original equation's solution is f(x) = 0.
From "WolframAlpha" reading:

https://www.wolframalpha.com/input/?...(f(x))%5E2%3D0

Is it correct what WolframAlpha says?

All the best,

Integrator

Last edited by Integrator; September 3rd, 2018 at 07:15 AM.
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