Question

y'''+4y'=e^x cos2x write the given differential equation in the form pf L(y)=g(x),where L is a linear differential operator with constant .IF possible , factor L.

Answer 1

$y''' + 4 y' = e^{x} cos(2x)$

$L(y) = g(x)$

$L(y) = \dfrac{d^{3}}{dx^{3}} + 4\dfrac{d}{dx}$

$g(x) = e^{x} cos(2x)$

Operator $L$ corresponds function $f(\tau) = \tau^{3} + 4\tau$.

$f(\tau) = \tau^{3} + 4\tau = \tau(\tau^2 + 4) = \tau(\tau + 2i)(\tau - 2i)$

Respectively,

$L(y) = \dfrac{d^{3}}{dx^{3}} + 4\dfrac{d}{dx} = \dfrac{d}{dx} \left(\dfrac{d^{2}}{dx^{2}} + 4 \right) = \dfrac{d}{dx} \left(\dfrac{d}{dx} + 2i \right)\left(\dfrac{d}{dx} - 2i \right)$