2
$\begingroup$

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 ?

Any advice greatly appreciated!

$\endgroup$
  • 3
    $\begingroup$ Can you try with $\sin x=\dfrac{e^{ix}-e^{-ix}}{2i}$. $\endgroup$ – Yadati Kiran Nov 15 '18 at 18:48
1
$\begingroup$

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

These are indeed complex exponentials.

$\endgroup$
2
$\begingroup$

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

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

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

Multiply out and get your equation out of it.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.