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The question is as above. I have not learnt about the Wronskian or the likes.

I suppose that I'll have to do something with differentiation or trying different x values.

EDIT:

  • $a$ and $b$ are non-zero and not equal to each other. I should have mentioned this earlier.

  • I have not learnt about the Wronskian formally and doubt that I should use it. It is nice to learn new methods and new theorems, however.

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    $\begingroup$ I assume that $a \neq b$ and both are non-zero? In that case suppose they are linearly dependent and show a contradiction. $\endgroup$
    – copper.hat
    Apr 16, 2015 at 0:30
  • $\begingroup$ A further hint: If they are linearly dependent, then there exists $c_1, c_2, c_3$ (with at least one of $c_i$ nonzero) such that $c_1 5 + c_2 e^{ax}+c_3e^{bx}=0$. $\endgroup$
    – JMoravitz
    Apr 16, 2015 at 0:40

3 Answers 3

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Let they be linearly dependent, i.e. $$ A\cdot5+Be^{ax}+Ce^{bx}\equiv0 $$ for $A$, $B$, $C$ not all equal to zero. Then it is true for example for $x=0$, 1, 2, i.e. $$ 5A+B+C=0, 5A+Be^a+Ce^b=0, 5A+Be^{2a}+Ce^{2b}=0. $$ The determinant is equal to $$ \begin{vmatrix} 5 & 1 &1\\ 5 & e^a &e^b\\ 5 & e^{2a} & e^{2b} \end{vmatrix}. $$ One can verify (please compute) that it in nonzero. But it means, that there is only one solution $A=B=C=0$. A contradiction.

BTW: Methods with wronskian are easier to use.

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(I'm assuming $a\ne b$, since otherwise they are dependent.)

One thing you might notice is that these all grow at different rates. A way to use that is the following:

If they are linearly dependent, we would have: $$A\cdot5+Be^{at}+Ce^{bt}=0$$ where the coefficients are nonzero.

WLOG, assume $a<b$. (If $a>b$, just switch around $a$ and $b$ in the following argument.) I'll also assume that they are both positive, but it's not hard to modify the argument for if they're negative.

Divide everything by $e^{bt}$ (we can do this since it's never equal to zero): $$5Ae^{-bt}+Be^{(a-b)t}+C=0$$ Taking the limit as $t\to\infty$, we have: $$C=0$$ which contradicts the coefficients being nonzero!

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  • $\begingroup$ where the coefficients are nonzero -> isn't this wrong? Linear dependence would mean that not all are zero, some may be. $\endgroup$
    – agdhruv
    Mar 25, 2018 at 20:34
  • $\begingroup$ You are absolutely correct. I was mistaken. $\endgroup$ Mar 25, 2018 at 20:38
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If any pair of these are linearly independent, then $cy_1 \neq y_2$, for any $c$. So, for example, $ce^{ax}\neq e^{bx}$. Suppose the opposite, $ce^{ax}=e^{bx}$. Dividing by $e^{ax}$, $c=e^{(b-a)x}$. But $c$ is a constant, it can't be a function of $x$. So we have a contradiction (if $a\neq b$), so they are linearly indpendent, and it works the same way if you use $5$ instead of one of the exponential functions.

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  • $\begingroup$ This proves that they are pairwise linearly independent, but doesn't yet prove they are mutually linearly independent. It requires a bit more effort to show that the three together can't combine somehow to equal zero. $\endgroup$
    – JMoravitz
    Apr 16, 2015 at 0:38

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