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 Differential Equations Ordinary and Partial Differential Equations Math Forum

 September 2nd, 2018, 07:08 AM #1 Member   Joined: Aug 2018 From: România Posts: 84 Thanks: 6 A differential equation Hello, Solve the differential equation $\displaystyle |f(x)-f'(x)|+(f(x))^2=0$. All the best, Integrator September 2nd, 2018, 07:12 AM #2 Senior Member   Joined: Sep 2016 From: USA Posts: 635 Thanks: 401 Math Focus: Dynamical systems, analytic function theory, numerics You need initial data to determine signs for $f,f'$. September 2nd, 2018, 07:28 AM #3 Global Moderator   Joined: Dec 2006 Posts: 20,919 Thanks: 2203 If f(x) is always real, f(x) = 0. September 2nd, 2018, 08:04 AM #4 Senior Member   Joined: Oct 2015 From: Greece Posts: 137 Thanks: 8 $\displaystyle |f(x) - \frac{df(x)}{dx}| + f^{2}(x) = 0$ 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 08:06 AM. September 2nd, 2018, 08:20 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, 09:33 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, 09:36 AM #7 Member   Joined: Aug 2018 From: România Posts: 84 Thanks: 6 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 09:49 AM. September 2nd, 2018, 09:43 AM #8 Global Moderator   Joined: Dec 2006 Posts: 20,919 Thanks: 2203 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, 11:34 AM #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, 12:07 PM #10 Senior Member   Joined: Oct 2015 From: Greece Posts: 137 Thanks: 8 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 12:10 PM. Tags differential, equation Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post jiasyuen Calculus 1 June 26th, 2016 02:04 PM Sonprelis Calculus 6 August 6th, 2014 10:07 AM interestedinmaths Differential Equations 2 January 22nd, 2014 02:15 AM PhizKid Differential Equations 0 February 24th, 2013 10:30 AM golomorf Differential Equations 4 August 6th, 2012 09:40 AM

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