September 2nd, 2018, 12:14 PM  #11 
Senior Member Joined: Oct 2015 From: Greece Posts: 137 Thanks: 8  
September 2nd, 2018, 01:04 PM  #12 
Senior Member Joined: Oct 2015 From: Greece Posts: 137 Thanks: 8 
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}{12}}(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}} $ 
September 2nd, 2018, 08:28 PM  #13 
Member Joined: Aug 2018 From: România Posts: 85 Thanks: 6 
"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 Last edited by skipjack; September 2nd, 2018 at 11:33 PM. 
September 2nd, 2018, 11:32 PM  #14 
Global Moderator Joined: Dec 2006 Posts: 20,927 Thanks: 2205 
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. 
September 3rd, 2018, 04:44 AM  #15  
Senior Member Joined: Oct 2015 From: Greece Posts: 137 Thanks: 8  Quote:
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 03:47 AM.  
September 3rd, 2018, 05:05 AM  #16 
Senior Member Joined: Oct 2015 From: Greece Posts: 137 Thanks: 8 
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 03:47 AM. 
September 3rd, 2018, 05:16 AM  #17 
Global Moderator Joined: Dec 2006 Posts: 20,927 Thanks: 2205 
As already posted, the original equation's solution is f(x) = 0.

September 3rd, 2018, 05:30 AM  #18  
Member Joined: Aug 2018 From: România Posts: 85 Thanks: 6  Quote:
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  
September 3rd, 2018, 05:42 AM  #19  
Member Joined: Aug 2018 From: România Posts: 85 Thanks: 6  Quote:
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 05:50 AM.  
September 3rd, 2018, 06:01 AM  #20  
Member Joined: Aug 2018 From: România Posts: 85 Thanks: 6  Quote:
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 06:15 AM.  

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