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

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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


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

= 0 = RHS

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