Question

In Problems 15–26 find a linear differential operator that annihilates

the given function.

$\mathbf{3}\mathbf{+}{\mathbf{e}}^{\mathbf{x}}\mathbf{cos}\mathbf{2}\mathbf{x}$

Short answer

$D\left(D^2-2D+5\right)$

Step-by-step solution

Annihilator Operator

If L is a linear differential operator with constant coefficients and f is a sufficiently differentiable function such that

$L\left(f\left(x\right)\right)=0$

then L is said to be an annihilator of the function

Using the Differential operators

We have the function,

3 + e^xcos 2x

And we have to annihilate it using a linear differential operator as the following technique:

xⁿis being annihilated by the linear differential operator Dⁿ⁺¹ , $e^{\alpha x}$is annihilated by $\left(D-\alpha \right)$and $\cos \beta x$is being annihilated by the linear differential operator $\left(D^{2^{}}+{\beta }^2\right)$, then $e^{\alpha x}{\cos }^{\beta x}$is being annihilated by the linear differential operator $\left(D^2-2\alpha D+{\beta }^2+{\alpha }^2\right)$then we annihilates the components of the given function as,

$3=3x^0$is annihilated by the linear differential operator , where n = 0.

$e^x\cos 2x$is annihilated by the linear differential operator

$\left(D^2-2D+2^2+1\right)=\left(D^2-2D+5\right)$, where $\alpha =1$and $\beta =2$.

Then we can annihilates the given function as

$D\left(D^2-2D+5\right)\left(3+e^x\cos 2x\right)=0$

Hence, the used differential operator islocalid="1667894536854" $\mathit{D}\mathbf{(}{\mathbf{D}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{D}\mathbf{+}\mathbf{5}\mathbf{)}$.