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

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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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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