# Finding the roots of a characteristic polynomial

Main aim is to find the lowest order equation with the solution: $$y(x)= 2 \cosh(x) + 3 e^{-2x} \sin(x)$$

Now, I am trying to find the roots to form the characteristic polynomial from which I get the lowest order equation.

However, am stuck with the second expression as the first can be easily expressed as $$e^x - e^{-x}$$ so I deduce $$\lambda_{1}=1,\lambda_{1} = -1$$ but the other expression I am not quite sure whether it is $$-2\pm i$$ or something else as there is an exponential and a trigonometric function at the same time ?

• Can you try with $\sin x=\dfrac{e^{ix}-e^{-ix}}{2i}$. – Yadati Kiran Nov 15 '18 at 18:48

$$2ie^{-2x}\sin x=e^{-2x}(e^{ix}-e^{-ix})=e^{(-2+i)x}-e^{(-2-i)x}.$$

These are indeed complex exponentials.

$$(\lambda -1)(\lambda +1)(\lambda +2-i)(\lambda +2+i)$$

$$= (\lambda ^2 -1)((\lambda+2)^2+1))$$

$$=(\lambda ^2 -1)(\lambda^2+4\lambda +5)$$

Multiply out and get your equation out of it.

• The OP is not asking the polynomial but how to handle the second term. – Yves Daoust Nov 15 '18 at 20:35