Answer
$\{e^x,e^{-3x}\}$
Work Step by Step
We are given:
$y''+2y'-3y=0$
Let $y=e^{ux}$
$y'=ue^{ux}$
$y''=u^2 e^{ux}$
The equation becomes:
$u^2e^{ux}+2ue^{ux}-3e^{ux}=0$
$(u^2+2u-3)e^{ux}=0$
$u^2+2u-3=0$
$(u-1)(u+3)=0$
$u=1$ or $u=-3$
Therefore $y=e^xc_1+e^{-3x}c_2$
Hence, the basis for the given equation is $\{e^x,e^{-3x}\}$
