September 2nd, 2018, 07:08 AM  #1 
Newbie Joined: Aug 2018 From: România Posts: 10 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, 07:12 AM  #2 
Senior Member Joined: Sep 2016 From: USA Posts: 444 Thanks: 254 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: 19,542 Thanks: 1751 
If f(x) is always real, f(x) = 0.

September 2nd, 2018, 08:04 AM  #4 
Member Joined: Oct 2015 From: Greece Posts: 78 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 08:06 AM. 
September 2nd, 2018, 08:20 AM  #5 
Newbie Joined: Aug 2018 From: România Posts: 10 Thanks: 2  
September 2nd, 2018, 09:33 AM  #6  
Newbie Joined: Aug 2018 From: România Posts: 10 Thanks: 2  Quote:
An idea: It is observed that $\displaystyle f (x) = g (e^{x+c})$... All the best, Integrator  
September 2nd, 2018, 09:36 AM  #7 
Newbie Joined: Aug 2018 From: România Posts: 10 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 09:49 AM. 
September 2nd, 2018, 09:43 AM  #8 
Global Moderator Joined: Dec 2006 Posts: 19,542 Thanks: 1751 
You're "overthinking" it. If f(x) is real, both terms are nonnegative, so they must both be zero if their sum is zero.

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 
Member Joined: Oct 2015 From: Greece Posts: 78 Thanks: 6 
JOmg yes!!! You mean $\displaystyle let: f(x) = u^{\frac{1}{1n}} $ Where n here is 2. Forgot that. Last edited by babaliaris; September 2nd, 2018 at 12:10 PM. 

Tags 
differential, equation 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
differential equation  jiasyuen  Calculus  1  June 26th, 2016 02:04 PM 
Gompertz equation  differential equation  Sonprelis  Calculus  6  August 6th, 2014 10:07 AM 
Differential equation  interestedinmaths  Differential Equations  2  January 22nd, 2014 02:15 AM 
Show that an equation satisfies a differential equation  PhizKid  Differential Equations  0  February 24th, 2013 10:30 AM 
differential equation  golomorf  Differential Equations  4  August 6th, 2012 09:40 AM 