**Given,**

y = Pe^{ax} + Qe^{bx}

**On differentiating with respect to x, we have**

\(\frac{dy}{dx}\) = Pae^{ax} + Qbe^{bx}

**Again, differentiating with respect to x, we have**

\(\frac{d^2y}{dx^2}\) = Pa^{2}e^{ax} + Qb^{2}e^{bx}

**Now,**

LHS = \(\frac{d^2y}{dx^2}\) - (a+b)\(\frac{dy}{dx}\)+ aby

= 0 **= RHS**