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Higher Order Derivatives

If y= aepx+...

Question

If $y = a e^{p x} + b e^{q x}$ , prove that $y_{2} - (p + q) y_{1} + p q y = 0$

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Solution

$y = a e^{p x} + b e^{q x}$

Differentiate on both sides w.r.t $x$

$\frac{d y}{d x} = y_{1} = a p e^{p x} + b q e^{q x}$

$\frac{d^{2} y}{d x^{2}} = y_{2} = a p^{2} e^{p x} + b q^{2} e^{q x}$

$y_{2} - (p + q) y_{1} + p q y = a p^{2} e^{p x} + b q^{2} e^{q x} - (p + q) (a p e^{p x} + b q e^{q x}) + p q (a e^{p x} + b e^{q x})$

$= 0$

$∴ y = y_{2} - (p + q) y_{1} + p q y = 0$

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