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I came across this question from my textbook, but in the text it only talked about $f ''+f =0$ when discussing trig functions. Now this questions at the end of the log and exp chapter with no discussion of it in this chapter. If anybody could please help.

It says: Find all functions f which are twice differentiable and satisfy the equation $f '' = f '$.

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Hint: When does $f' = f$? –  JavaMan Feb 21 '13 at 8:27
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Take $u=f'$. The equation is $u=u'$. –  Pedro Tamaroff Feb 21 '13 at 8:34
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4 Answers

up vote 6 down vote accepted

Suppose that $f'' \equiv 0$. Then, we have that

$$f' = 0 \implies f(x) = c$$

for some constant $c$. If $f \not \equiv 0$ then rewrite your relation as:

$$\displaystyle \frac{f'}{f''} = 1$$

Integrating both sides, you obtain:

$$\displaystyle \int \frac{f'}{f''} = \int 1 \implies \log(|f'|) = x + c \implies |f'| = e^{x+c} \implies f' = e^{x+c}$$ since $e^{x+c} > 0 \ \forall x \in \mathbb{R}$. Note that $c \in \mathbb{R}$ is some arbitrary constant. You can in fact rewrite the above as

$$f' = Ce^x \ \ \text{where} \ C = e^c$$ Re-integrating:

$$\int f' = C\int e^x \implies f = Ce^x + a$$ So any function satisfying the given differential equation must be of the form $f(x) = Ce^x + a$

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You have missed out a constant at the last integration step. –  Abhra Abir Kundu Feb 21 '13 at 8:40
    
Good it looks the same now math.stackexchange.com/a/309977/48639 –  Abhra Abir Kundu Feb 21 '13 at 8:44
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Thank you for pointing that out! Thanks @Chris Taylor for editing the solution. –  Orest Xherija Feb 21 '13 at 8:44
    
That's impossible! The first derivative will be some number but the second will be 0. How are they working? –  Orest Xherija Feb 21 '13 at 8:49
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Thanks to everybody. –  Dick Feb 21 '13 at 8:52
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The ODE is a homogeneous linear OE with constant coefficients. First set the equation $m^2-m=0$ and then the general solution will be $$y_c=C_1x^0+C_2\exp(x)$$

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Short and sweet! +1 –  amWhy Feb 21 '13 at 13:19
    
@amWhy: Mornin. –  B. S. Feb 21 '13 at 13:19
    
Isn't this assuming that @Dick is aware of the theory of differential equations? It seems to me that this is merely calculus. –  user44069 Feb 22 '13 at 0:53
    
@Stefan: Dear Stefan, when he tagged the DE, I assumed he knew that. I really don't know what to do in this case? Should I ask the OP with longs comment if he/she is aware of this text or that source or...This is really exhausting here, anyway. :-) –  B. S. Feb 22 '13 at 6:31
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From this it implies,

$$\frac{df'(x)}{dx}=f'(x)$$

$$\Rightarrow \frac{df'(x)}{f'(x)}=dx$$

Integrating we have,

$$\ln(f'(x))=x+c$$

$$\Rightarrow f'(x)=e^x.e^c$$

$$\Rightarrow c_1(\frac{df(x)}{dx})=e^x$$

$c_1=1/e^c$

$$c_1d(f(x))=e^xdx$$

Integrating we have,

$$c_1f(x)=e^x+c_2$$

$$\Rightarrow f(x)=c_3e^x+c_2$$

$c,c_1,c_2,c_3$ are integration constant.

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@Chris Taylor and @Orest Xherija I think that the proof is not rigorous since we can't write $$\frac{f'}{f''}=1$$ if we do not suppose that $f''$ does not vanish. To avoid this problem and without using the characteristic polynomial of differential equation I suggest:

We pose $u=f'$ and $v=e^{-x}u$, so we can easily verify that $v'=0$ and that $v$ is a constant say $c_1$, and we conclude that $u=c_1 e^x$ and with integration $f=c_1 e^x+ c_2$.

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You are indeed correct to point out that flaw. But that does not mean the proof is not rigorous. It only means that there is one more case to be checked. I will add it. –  Orest Xherija Feb 22 '13 at 0:48
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