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September 2nd, 2018, 01:14 PM   #11
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Quote:
 Originally Posted by Integrator 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
c will be cancelled if you multiply both sides of the equation with μ0.

 September 2nd, 2018, 02:04 PM #12 Member   Joined: Oct 2015 From: Greece Posts: 81 Thanks: 6 So back to the problem : $\displaystyle f(x) - \frac{df(x)}{dx} + f^{2}(x) = 0 \Leftrightarrow -\frac{df(x)}{dx} + f(x) = -f^{2}(x)$ $\displaystyle let:f(x) = u^{\frac{1}{1-2}}(x) = u^{-1}(x)$ $\displaystyle -\frac{du^{-1}(x)}{dx} + u^{-1}(x) = -u^{-2}(x) \Leftrightarrow u^{-2}(x) \cdot \frac{du(x)}{dx} + u^{-1}(x) = -u^{-2}(x) \Leftrightarrow \frac{u^{-2}(x)}{u^{-2}(x)} \cdot \frac{du(x)}{dx} + \frac{u^{-1}(x)}{u^{-2}(x)} = \frac{-u^{-2}(x)}{u^{-2}(x)} \Leftrightarrow \frac{du(x)}{dx} + u(x) = -1$ $\displaystyle let:μ_{0} = e^{\int 1dx} = e^{x+c}$ $\displaystyle \frac{du(x)}{dx} \cdot e^{x} + u(x) \cdot e^{x} = -e^{x} \Leftrightarrow \frac{d[u(x) \cdot e^{x}]}{dx} = -e^{x} \Leftrightarrow \int \frac{d[u(x) \cdot e^{x}]}{dx} \cdot dx = \int -e^{x} dx \Leftrightarrow u(x) \cdot e^{x} + c_{1} = \int -e^{x} dx \Leftrightarrow u(x) \cdot e^{x} + c_{1} = -e^{x} - c_{2} \Leftrightarrow u(x) = \frac{-e^{x}}{e^{x}} - \frac{c_{2} + c_{1}}{e^{x}} \Leftrightarrow u(x) = -1 - ke^{-x} : where : k = (c_{2} + c_{1})$ $\displaystyle f(x) = u^{-1}(x) \Leftrightarrow f(x) = (-1 - ke^{-x})^{-1}$ Finally: $\displaystyle f(x) = \frac{1}{-1 - ke^{-x}}$
 September 2nd, 2018, 09:28 PM #13 Newbie   Joined: Aug 2018 From: România Posts: 24 Thanks: 2 "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 Thanks from babaliaris Last edited by skipjack; September 3rd, 2018 at 12:33 AM.
 September 3rd, 2018, 12:32 AM #14 Global Moderator   Joined: Dec 2006 Posts: 20,281 Thanks: 1965 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. Thanks from babaliaris
September 3rd, 2018, 05:44 AM   #15
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Quote:
 Originally Posted by skipjack 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.
Well this shows you have more experience.

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 04:47 AM.

 September 3rd, 2018, 06:05 AM #16 Member   Joined: Oct 2015 From: Greece Posts: 81 Thanks: 6 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 04:47 AM.
 September 3rd, 2018, 06:16 AM #17 Global Moderator   Joined: Dec 2006 Posts: 20,281 Thanks: 1965 As already posted, the original equation's solution is f(x) = 0. Thanks from babaliaris
September 3rd, 2018, 06:30 AM   #18
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Quote:
 Originally Posted by skipjack 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.
Hello,

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, 06:42 AM   #19
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Quote:
 Originally Posted by babaliaris So back to the problem : $\displaystyle f(x) - \frac{df(x)}{dx} + f^{2}(x) = 0 \Leftrightarrow -\frac{df(x)}{dx} + f(x) = -f^{2}(x)$ $\displaystyle let:f(x) = u^{\frac{1}{1-2}}(x) = u^{-1}(x)$ $\displaystyle -\frac{du^{-1}(x)}{dx} + u^{-1}(x) = -u^{-2}(x) \Leftrightarrow u^{-2}(x) \cdot \frac{du(x)}{dx} + u^{-1}(x) = -u^{-2}(x) \Leftrightarrow \frac{u^{-2}(x)}{u^{-2}(x)} \cdot \frac{du(x)}{dx} + \frac{u^{-1}(x)}{u^{-2}(x)} = \frac{-u^{-2}(x)}{u^{-2}(x)} \Leftrightarrow \frac{du(x)}{dx} + u(x) = -1$ $\displaystyle let:μ_{0} = e^{\int 1dx} = e^{x+c}$ $\displaystyle \frac{du(x)}{dx} \cdot e^{x} + u(x) \cdot e^{x} = -e^{x} \Leftrightarrow \frac{d[u(x) \cdot e^{x}]}{dx} = -e^{x} \Leftrightarrow \int \frac{d[u(x) \cdot e^{x}]}{dx} \cdot dx = \int -e^{x} dx \Leftrightarrow u(x) \cdot e^{x} + c_{1} = \int -e^{x} dx \Leftrightarrow u(x) \cdot e^{x} + c_{1} = -e^{x} - c_{2} \Leftrightarrow u(x) = \frac{-e^{x}}{e^{x}} - \frac{c_{2} + c_{1}}{e^{x}} \Leftrightarrow u(x) = -1 - ke^{-x} : where : k = (c_{2} + c_{1})$ $\displaystyle f(x) = u^{-1}(x) \Leftrightarrow f(x) = (-1 - ke^{-x})^{-1}$ Finally: $\displaystyle f(x) = \frac{1}{-1 - ke^{-x}}$
Hello,

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 06:50 AM.

September 3rd, 2018, 07:01 AM   #20
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Quote:
 Originally Posted by skipjack As already posted, the original equation's solution is f(x) = 0.

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 07:15 AM.

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