Question

show that the set of fuctions y1=e^-2x , y2=e^6x is linearly independent on (minus infinity, positive infinity)

Answer 1

Solution:

Consider $\small ay_{1}+by_{2}=0$

$\small ae^{-2x}+be^{6x}=0$

Put $\small x=0$

$\small \therefore a+b=0.........\left ( i \right )$

Differentiate $\small ae^{-2x}+be^{6x}=0$    with respect to $\small x$ .

$\small \therefore -2ae^{-2x}+6be^{6x}=0$

Put $\small x=0$

$\small \therefore -2a+6b=0\Rightarrow -a+3b=0...............\left ( ii \right )$

Solving $\small \left ( i \right )$    and $\small \left ( ii \right )$    simultaneously we get,

$\small a=b=0$

Thus, $\small y_{1}$    and $\small y_{2}$ are linearly independent   on $\small \left ( -\infty ,\infty \right )$ .