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Question
If y = ae2x + be−x, show that, \[\frac{d^2 y}{d x^2} - \frac{dy}{dx} - 2y = 0\] ?
Solution
Here,
\[y = a e^{2x} + b e^{- x} \]
\[\text { Differentiating w . r . t . x, we get } \]
\[\frac{d y}{d x} = 2a e^{2x} - b e^{- x} \]
\[\text { Differentiating again w . r . t . x, we get }\]
\[\frac{d^2 y}{d x^2} = 4a e^{2x} + b e^{- x} \]
\[ \Rightarrow \frac{d^2 y}{d x^2} = 2a e^{2x} - b e^{- x} + 2\left( a e^{2x} + b e^{- x} \right)\]
\[ \Rightarrow \frac{d^2 y}{d x^2} = \frac{d y}{d x} + 2y \]
\[ \Rightarrow \frac{d^2 y}{d x^2} - \frac{d y}{d x} - 2y = 0\]
Hence proved.
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