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September 2nd, 2018, 08:08 AM   #1
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A differential equation

Hello,

Solve the differential equation $\displaystyle |f(x)-f'(x)|+(f(x))^2=0$.

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Integrator
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September 2nd, 2018, 08:12 AM   #2
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You need initial data to determine signs for $f,f'$.
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September 2nd, 2018, 08:28 AM   #3
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If f(x) is always real, f(x) = 0.
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September 2nd, 2018, 09:04 AM   #4
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$\displaystyle
|f(x) - \frac{df(x)}{dx}| + f^{2}(x) = 0
$

If:
$\displaystyle
f(x) = \frac{df(x)}{dx}
$

Then :

$\displaystyle
0 + f^{2}(x) = 0 \Leftrightarrow f(x) = 0
$

Else if:
$\displaystyle
f(x) > \frac{df(x)}{dx}
$

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

$\displaystyle
μ_{0} = e^{\int -1dx} = e^{-1 \cdot \int dx} \Leftrightarrow
μ_{0} = e^{-x + c}
$

$\displaystyle
\frac{df(x)}{dx} \cdot e^{-x} - f(x) \cdot e^{-x} = f^{2}(x) \cdot e^{-x}
\Leftrightarrow
\frac{d[f(x) \cdot e^{-x}]}{dx} = f^{2}(x) \cdot e^{-x}
\Leftrightarrow
\int \frac{d[f(x) \cdot e^{-x}]}{dx} \cdot dx = \int f^{2}(x) \cdot e^{-x} \cdot dx
\Leftrightarrow
f(x) \cdot e^{-x} + c_{1} = \int f^{2}(x) \cdot e^{-x}
$



Well of course here I'm stuck. Any Ideas?

Last edited by babaliaris; September 2nd, 2018 at 09:06 AM.
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September 2nd, 2018, 09:20 AM   #5
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Quote:
Originally Posted by skipjack View Post
If f(x) is always real, f(x) = 0.
$\displaystyle f (x) = 0$ is the common solution.
However, some say that applying the module definition also leads to two other function families.

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September 2nd, 2018, 10:33 AM   #6
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Quote:
Originally Posted by babaliaris View Post
$\displaystyle
|f(x) - \frac{df(x)}{dx}| + f^{2}(x) = 0
$

If:
$\displaystyle
f(x) = \frac{df(x)}{dx}
$

Then :

$\displaystyle
0 + f^{2}(x) = 0 \Leftrightarrow f(x) = 0
$

Else if:
$\displaystyle
f(x) > \frac{df(x)}{dx}
$

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

$\displaystyle
μ_{0} = e^{\int -1dx} = e^{-1 \cdot \int dx} \Leftrightarrow
μ_{0} = e^{-x + c}
$

$\displaystyle
\frac{df(x)}{dx} \cdot e^{-x} - f(x) \cdot e^{-x} = f^{2}(x) \cdot e^{-x}
\Leftrightarrow
\frac{d[f(x) \cdot e^{-x}]}{dx} = f^{2}(x) \cdot e^{-x}
\Leftrightarrow
\int \frac{d[f(x) \cdot e^{-x}]}{dx} \cdot dx = \int f^{2}(x) \cdot e^{-x} \cdot dx
\Leftrightarrow
f(x) \cdot e^{-x} + c_{1} = \int f^{2}(x) \cdot e^{-x}
$



Well of course here I'm stuck. Any Ideas?
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
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September 2nd, 2018, 10:36 AM   #7
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babaliaris,

I think that $\displaystyle μ_{0} = e^{x + c}$.
An idea:
It is observed that $\displaystyle f (x) = g (e^{x+c})=\frac{a_1e^{x+c}+a_2}{a_3e^{x+c}+a_4}$ where $\displaystyle a_i$ are constants to be found...

All the best,

Integrator

Last edited by Integrator; September 2nd, 2018 at 10:49 AM.
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September 2nd, 2018, 10:43 AM   #8
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You're "overthinking" it. If f(x) is real, both terms are non-negative, so they must both be zero if their sum is zero.
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September 2nd, 2018, 12:34 PM   #9
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Have you discussed Bernoulli's equation in your Differential Equations course? After separating the absolute value, you get two equations that involve a nonlinear term.

You can use a "change of variables" approach to reduce the order to a linear equation, and then solve from there. This particular equation will have a power of 2, which will yield the "logistic equation" as the solution.

If any of this sounds familiar from your coursework, please feel free to let me know. If you are still unable to solve it, then I don't mind typing it up for you.

Note: If/when you do solve this problem, it will be necessary to check your answer, as an extraneous function may or may not be found.

Hope that helps!
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September 2nd, 2018, 01:07 PM   #10
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JOmg yes!!! You mean
$\displaystyle
let:
f(x) = u^{\frac{1}{1-n}}
$

Where n here is 2.

Forgot that.

Last edited by babaliaris; September 2nd, 2018 at 01:10 PM.
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