Question

If $y=e^{4x}+2e^{-x}$ satisfies the equation $y_3+A{y}_1+By=0$ then the value of $AB$ is

Answer 1

$y_1=4e^{4x}-2e^{-x}\Rightarrow y_2=16e^{4x}+2e^{-x}\Rightarrow y_3=64e^{4x}-2e^{-x}$
Putting these values in $y_3+A{y}_1+By=0$, we have
$64e^{4x}-2e^{-x}+A(4e^{4x}-2e^{-x})+B(e^{4x}+2e^{-x})=0$
$(64+4A+B)e^{4x}+(-2-2A+2B)e^{-x}=0$
This is true for all x only if $64+4A+B=0$ and $2+2A-2B=0$. Solving we get $A=-13$ and $B=-12$. i.e. $AB=156$.