Given,
y = Peax + Qebx
On differentiating with respect to x, we have
\(\frac{dy}{dx}\) = Paeax + Qbebx
Again, differentiating with respect to x, we have
\(\frac{d^2y}{dx^2}\) = Pa2eax + Qb2ebx
Now,
LHS = \(\frac{d^2y}{dx^2}\) - (a+b)\(\frac{dy}{dx}\)+ aby
= 0 = RHS