Question

Show that y = be^x + ce^2x is a solution of the differential equation, \[\frac{d^2 y}{d x^2} - 3\frac{dy}{dx} + 2y = 0\]

Answer 1

We have,

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

Differentiating both sides of equation (1) with respect to x, we get

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

Differentiating both sides of equation (2) with respect to x, we get

\[\frac{d^2 y}{d x^2} = b e^x + 4c e^{2x} \]

\[ = 3b e^x + 6c e^{2x} - 2b e^x - 2c e^{2x} \]

\[ = 3\left( b e^x + 2c e^{2x} \right) - 2\left( b e^x + c e^{2x} \right)\]

\[ = 3\frac{dy}{dx} - 2y ..........\left[\text{Using equations }\left( 1 \right)\text{ and }\left( 2 \right) \right]\]

\[\Rightarrow \frac{d^2 y}{d x^2} - 3\frac{dy}{dx} + 2y = 0\]

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