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Are the functions $f, g, h$ given below linearly independent? If they are not linearly independent, find a nontrivial solutions to the equations below:

$$f(x)=e^{2x}- \cos(9x), \quad g(x)=e^{2x}+ \cos(9x), \quad h(x)= \cos(9x)$$

My take so far is that, I know these functions are not linearly independent (meaning this is surely dependent) by Wronskian (as the Wronskian determinant isn't equal to zero), but I have no idea how to find nontrivial solutions to this question. I think I should have an answer as follows:

$$ C_1(e^{2x}-\cos(9x)) + C_2(e^{2x}+\cos(9x)) + C_3\cos(9x) = 0 \qquad \text{ where } Cs \text{ are constant } $$

Could I get some help on finding those constants? I've tried $C_1$ as $-1$, $C_2$ as $1$ and $C_3$ as $0$ but that surely cannot be the answer.

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  • $\begingroup$ In Mathematica, Wronskian[{Exp[2 x] - Cos[9 x], Exp[2 x] + Cos[9 x], Cos[9 x]}, x] yields $0$, showing that the three functions are linearly dependent. $\endgroup$ – David G. Stork Feb 8 '16 at 18:19
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Those functions are a red herring. ;-)

You basically have the vectors $$ v-w,\quad v+w,\quad w $$ in some vector space. These vectors belong to the subspace spanned by $v$ and $w$, which has dimension at most $2$. So a set of three vectors is necessarily linearly dependent.

How to find a nonzero linear combination is easy: $$ a(v-w)+b(v+w)+cw=0 $$ gives $$ (a+b)v+(-a+b+c)w=0 $$ so we can choose $b=-a$ and so $c=2a$. If we take $a=1$, we obtain $b=-1$ and $c=2$.

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You can take $C_1 = 1, C_2 = -1$ and $ C_3 = -2$.

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  • $\begingroup$ may i ask how you could find those answers? $\endgroup$ – Allie Feb 8 '16 at 18:05
  • $\begingroup$ I choose $C_1 = 1$ (without any reason). Then I wanted to cancel the two exponential so I had to choose $C_2 = -1$. Then I was left with $-2\cos(9x)$ therefore I had to pick $C_3=-2$. $\endgroup$ – Onil90 Feb 8 '16 at 18:10
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$$\begin{bmatrix} f(x)\\ g(x)\\ h(x) \end{bmatrix}= \begin{bmatrix} 1 & -1\\ 1 & 1 \\ 0 & 1 \\ \end{bmatrix}\cdot \begin{bmatrix} -e^{2x}\\ \cos(9x) \end{bmatrix}$$

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