Question

Show that y = ae^2x + be^−x is a solution of the differential equation \[\frac{d^2 y}{d x^2} - \frac{dy}{dx} - 2y = 0\]

Answer 1

We have,

\[y = a e^{2x} + b e^{- x}...........(1)\]

Differentiating both sides of equation (1) with respect to

`x,` we get

\[\frac{dy}{dx} = 2a e^{2x} - b e^{- x}..........(2)\]

Differentiating both sides of equation (2) with respect to

`x,` we get

\[\frac{d^2 y}{d x^2} = 4a e^{2x} + b e^{- x} \]
\[ \Rightarrow \frac{d^2 y}{d x^2} = 2a e^{2x} - b e^{- x} + 2a e^{2x} + 2b e^{- x} \]
\[ \Rightarrow \frac{d^2 y}{d x^2} = \left( 2a e^{2x} - b e^{- x} \right) + 2\left( a e^{2x} + b e^{- x} \right)\]
\[ \Rightarrow \frac{d^2 y}{d x^2} = \frac{dy}{dx} + 2y ..........\left[\text{Using equations (1) and (2)} \right]\]

\[\Rightarrow\frac{d^2 y}{d x^2} - \frac{dy}{dx} - 2y = 0\]

Hence, the given function is the solution to the given differential equation.