I must find an ecuation like f(x) and to resolve this. f(x)+f`(x)=0 . (where f`(x) is derivative) I made smth but it`s correct. my teacher said to me that f(x) is smth like sinx*e^x but with something in addition. Thnx
$\displaystyle \textup{f(x)}+\textup{f'(x)}=0\Rightarrow e^x\left ( \textup{f(x)}+\textup{f'(x)} \right )=0\Rightarrow$
$\displaystyle \Rightarrow e^x\textup{f(x)}+e^x\textup{f'(x)}=0 \Leftrightarrow $
$\displaystyle \Leftrightarrow \left (e^x \right )'\textup{f(x)}+e^x\textup{f'(x)}=0 \Leftrightarrow $
$\displaystyle \Leftrightarrow \left ( e^x \textup{f(x)}\right )'=0$
$\displaystyle \left ( e^x \textup{f(x)}\right )'=0\Rightarrow $
$\displaystyle \Rightarrow e^x\textup{f(x)}=c,\, c \in \mathbb{R} \Rightarrow \textup{f(x)}=ce^{-x}$
What you are saying is that you want to solve the differential equation y''''+ y= 0. That is a linear equation with constant coefficients. If you try a solution of the form $\displaystyle y= e^{rx}$, the fourth derivative is $\displaystyle r^4e^{rx}$ so the equation becomes $\displaystyle r^4e^{rx}+ e^{rx}= e^x(r^4+ 1)= 0$. That's the "characteristic equation" for this differential equation. It is equivalent to $\displaystyle x^4= -1$ which has four complex solutions, two pairs of complex conjugates. To write the solution to the differential equation in terms of real functions, use [tex]e^{a+ bi}= e^ae^{bi}= e^a(cos(b)+ i sin(b))
I'll call f(x) "y", just because I prefer Leibnitz notation to Newtonian.
$\displaystyle \displaystyle \begin{align*} y + \frac{dy}{dx} &= 0 \\ \frac{dy}{dx} = -y \\ \frac{1}{y}\,\frac{dy}{dx} &= -1 \\ \int{\frac{1}{y}\,\frac{dy}{dx}\,dx} &= \int{-1\,dx} \\ \int{\frac{1}{y}\,dy} &= -x + C_1 \\ \ln{|y|} + C_2 &= -x + C_1 \\ \ln{|y|} &= -x + C_1 - C_2 \\ |y| &= e^{-x + C_1 - C_2} \\ |y| &= e^{C_1 - C_2}\,e^{-x} \\ y &= A\,e^{-x} \textrm{ where } A = \pm e^{C_1 - C_2} \end{align*}$
There you go, all functions of the form $\displaystyle \displaystyle \begin{align*} f(x) = A\,e^{-x} \end{align*}$ will satisfy $\displaystyle \displaystyle \begin{align*} f(x) + f'(x) = 0 \end{align*}$.