If $y = a e^{4 x} + b e^{5 x}$ , then prove $y_{2} - 9 y_{1} + 20 y = 0$

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Solution

$y_{1} = \frac{d y}{d x} = 4 a e^{4 x} + 5 b e^{5 x}$

$y_{2} = \frac{d^{2} y}{d x^{2}} = 16 a e^{4 x} + 25 b e^{5 x}$

$\Rightarrow y_{2} - 9 y_{1} + 20 y = 16 a e^{4 x} + 25 b e^{5 x} - 9 (4 a e^{4 x} + 5 b e^{5 x}) + 20 (a e^{4 x} + b e^{5 x})$

$= - 20 a e^{4 x} + 20 a e^{4 x} - 20 b e^{5 x} + 20 b e^{5 x} = 0$

Hence proved

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