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 Differential Equations Ordinary and Partial Differential Equations Math Forum

September 2nd, 2018, 12:14 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, 01:04 PM #12 Senior Member   Joined: Oct 2015 From: Greece Posts: 137 Thanks: 8 So back to the problem : $\displaystyle f(x) - \frac{df(x)}{dx} + f^{2}(x) = 0 \Leftrightarrow -\frac{df(x)}{dx} + f(x) = -f^{2}(x)$ $\displaystyle let:f(x) = u^{\frac{1}{1-2}}(x) = u^{-1}(x)$ $\displaystyle -\frac{du^{-1}(x)}{dx} + u^{-1}(x) = -u^{-2}(x) \Leftrightarrow u^{-2}(x) \cdot \frac{du(x)}{dx} + u^{-1}(x) = -u^{-2}(x) \Leftrightarrow \frac{u^{-2}(x)}{u^{-2}(x)} \cdot \frac{du(x)}{dx} + \frac{u^{-1}(x)}{u^{-2}(x)} = \frac{-u^{-2}(x)}{u^{-2}(x)} \Leftrightarrow \frac{du(x)}{dx} + u(x) = -1$ $\displaystyle let:μ_{0} = e^{\int 1dx} = e^{x+c}$ $\displaystyle \frac{du(x)}{dx} \cdot e^{x} + u(x) \cdot e^{x} = -e^{x} \Leftrightarrow \frac{d[u(x) \cdot e^{x}]}{dx} = -e^{x} \Leftrightarrow \int \frac{d[u(x) \cdot e^{x}]}{dx} \cdot dx = \int -e^{x} dx \Leftrightarrow u(x) \cdot e^{x} + c_{1} = \int -e^{x} dx \Leftrightarrow u(x) \cdot e^{x} + c_{1} = -e^{x} - c_{2} \Leftrightarrow u(x) = \frac{-e^{x}}{e^{x}} - \frac{c_{2} + c_{1}}{e^{x}} \Leftrightarrow u(x) = -1 - ke^{-x} : where : k = (c_{2} + c_{1})$ $\displaystyle f(x) = u^{-1}(x) \Leftrightarrow f(x) = (-1 - ke^{-x})^{-1}$ Finally: $\displaystyle f(x) = \frac{1}{-1 - ke^{-x}}$ September 2nd, 2018, 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 Thanks from babaliaris 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. Thanks from babaliaris September 3rd, 2018, 04:44 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 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. Thanks from babaliaris September 3rd, 2018, 05:30 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, 05:42 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 05:50 AM. September 3rd, 2018, 06:01 AM   #20
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Quote:
 Originally Posted by skipjack As already posted, the original equation's solution is f(x) = 0.
From "WolframAlpha" reading:

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. 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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