Question

If y = ae^2x + be^−x, show that, \[\frac{d^2 y}{d x^2} - \frac{dy}{dx} - 2y = 0\] ?

Answer 1

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.