Question

How am I supposed to do this question​

Answer 1

$\: \: \: \: \: \: \: \: \: \: \: \: y = 2 {e}^{3x} + 5 {e}^{4x} = > 12y = 24 {e}^{3x} + 60 {e}^{4x} \\ \: \: \: \: \: \: \: \: \: \frac{dy}{dx} = \frac{d(2 {e}^{3x}) }{dx} + \frac{d(5 {e}^{4x}) }{dx} \\ \: \: \: \: \: \: \: \: \: \frac{dy}{dx} = 2.3 {e}^{3x} + 5.4 {e}^{4x} \: \: \: \: \: \: ...( \frac{d( {e}^{ax} )}{dx} = a {e}^{ax}) \\ \: \: \: \: \: \: \: \: \: \frac{dy}{dx} = 6 {e}^{3x} + 20 {e}^{4x} \\ \: \: \: \: \: \: \: 7 \frac{dy}{dx} = 42 {e}^{3x} + 140 {e}^{4x} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ... \: (i) \\ \frac{d}{dx} ( \frac{dy}{dx} ) = \frac{d(6 {e}^{3x}) }{dx} + \frac{d( {20e}^{4x} )}{dx} \\ \: \: \: \: \: \: \: \frac{ {d}^{2}y }{d {x}^{2} } = 6.3 {e}^{3x} + 20.4 {e}^{4x} \\ \: \: \: \: \: \: \: \frac{ {d}^{2}y }{d {x}^{2} } = {18e}^{3x} + {80e}^{4x} \: \: \: \: \: \: \: \: \: \: \: ... \: (ii) \\ \\ \: \: \: \: \: \frac{ {d}^{2} y}{dx} - 7 \frac{dy}{dx} + 12y \\ = 18 {e}^{3x} + 80 {e}^{4x} - (42 {e}^{3x} + 140 {e}^{4x} ) + {24e}^{3x} + {60e}^{4x} \\ = 18 {e}^{3x} + 80 {e}^{4x} - 42 {e}^{3x} - 140 {e}^{4x} + 24 {e}^{3x} + 60 {e}^{4x} \\ = 42 {e}^{3x } - {42e}^{3x} + 140 {e}^{4x} - {140e}^{4x} \\ = 0$

Hence, $\frac{{d}^{2}y}{dx} - 7 \frac{dy}{dx} + 12y$ = 0

... Hence Proved!