If y = Pe^ax + QeTbx, then show that d^2y/dx^2 - (a+b)dy/dx+ aby = 0 ​

Question

If y = Pe^ax + Qe^bx, then show that \(\frac{d^2y}{dx^2}\) - (a+b)\(\frac{dy}{dx}\)+ aby = 0

Answer 1

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

\(​​\frac{d^2y}{dx^2}\) =  Pa²e^ax + Qb²e^bx

Now,

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

= 0 = RHS