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 September 2nd, 2018, 08:08 AM #1 Newbie   Joined: Aug 2018 From: România Posts: 24 Thanks: 2 A differential equation Hello, Solve the differential equation $\displaystyle |f(x)-f'(x)|+(f(x))^2=0$. All the best, Integrator
 September 2nd, 2018, 08:12 AM #2 Senior Member   Joined: Sep 2016 From: USA Posts: 557 Thanks: 322 Math Focus: Dynamical systems, analytic function theory, numerics You need initial data to determine signs for $f,f'$.
 September 2nd, 2018, 08:28 AM #3 Global Moderator   Joined: Dec 2006 Posts: 20,281 Thanks: 1965 If f(x) is always real, f(x) = 0.
 September 2nd, 2018, 09:04 AM #4 Member   Joined: Oct 2015 From: Greece Posts: 81 Thanks: 6 $\displaystyle |f(x) - \frac{df(x)}{dx}| + f^{2}(x) = 0$ 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.
September 2nd, 2018, 09:20 AM   #5
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Quote:
 Originally Posted by skipjack 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.

All the best,

Integrator

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

 September 2nd, 2018, 10:36 AM #7 Newbie   Joined: Aug 2018 From: România Posts: 24 Thanks: 2 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.
 September 2nd, 2018, 10:43 AM #8 Global Moderator   Joined: Dec 2006 Posts: 20,281 Thanks: 1965 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. Thanks from v8archie
 September 2nd, 2018, 12:34 PM #9 Newbie   Joined: Dec 2016 From: Austin Posts: 16 Thanks: 1 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!
 September 2nd, 2018, 01:07 PM #10 Member   Joined: Oct 2015 From: Greece Posts: 81 Thanks: 6 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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