Answer
Given below.
Work Step by Step
On derivating the equation
$y=e^(ax)\cdot(c_1+c_2x)$
$y'=ae^(ax)\cdot(c_1+c_2x)+e^(ax)\cdot(0+c_2)$
$y'=(e^(ax))[a\cdot(c_1+c_2x)+(c_2)]$
On again derivating
$y''=ae^(ax)[a\cdot(c_1+c_2x)+(c_2)]+e^(ax)[a\cdot(0+c_2)+(0)]$
$y''=(e^(ax))\cdot[ac_1+c_2x+ac_2]$
On putting these valves in equation
$y''-2ay'+a^2y=0$ , we get
$(e^(ax))\cdot[ac_1+c_2x+ac_2]-2a(e^(ax))[a\cdot(c_1+c_2x)+(c_2)]+a^2e^(ax)\cdot(c_1+c_2x)=0$
$(e^(ax))\cdot[ac_1+c_2x+ac_2-2a^2(c_1+c_2x)-2ac_2+a^2(c_1+c_2x)=0$
$0=0$
Hence it is true for all real values of $x$
