derivatives

**thealivision** · March 14th 2013, 07:52 AM

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

**veileen** · March 14th 2013, 08:14 AM

$\textup{f(x)}+\textup{f'(x)}=0\Rightarrow e^x\left ( \textup{f(x)}+\textup{f'(x)} \right )=0\Rightarrow$

$\Rightarrow e^x\textup{f(x)}+e^x\textup{f'(x)}=0 \Leftrightarrow$

$\Leftrightarrow \left (e^x \right )'\textup{f(x)}+e^x\textup{f'(x)}=0 \Leftrightarrow$

$\Leftrightarrow \left ( e^x \textup{f(x)}\right )'=0$

$\left ( e^x \textup{f(x)}\right )'=0\Rightarrow$

$\Rightarrow e^x\textup{f(x)}=c,\, c \in \mathbb{R} \Rightarrow \textup{f(x)}=ce^{-x}$

**thealivision** · March 14th 2013, 08:18 AM

thnx veileen. but now i realise i wrong. the exercise was f(x)+f````(x)=0 where f````(x) is derivative by four times. thnx anyway

**HallsofIvy** · March 14th 2013, 08:54 AM

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 $y= e^{rx}$ , the fourth derivative is $r^4e^{rx}$ so the equation becomes $r^4e^{rx}+ e^{rx}= e^x(r^4+ 1)= 0$ . That's the "characteristic equation" for this differential equation. It is equivalent to $x^4= -1$ 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))

**Prove It** · March 14th 2013, 03:09 PM

Originally Posted by thealivision

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

I'll call f(x) "y", just because I prefer Leibnitz notation to Newtonian.

$\displaystyle \begin{align*} y + \frac{dy}{dx} &= 0 \\ \frac{dy}{dx} = -y \\ \frac{1}{y}\,\frac{dy}{dx} &= -1 \\ \int{\frac{1}{y}\,\frac{dy}{dx}\,dx} &= \int{-1\,dx} \\ \int{\frac{1}{y}\,dy} &= -x + C_1 \\ \ln{|y|} + C_2 &= -x + C_1 \\ \ln{|y|} &= -x + C_1 - C_2 \\ |y| &= e^{-x + C_1 - C_2} \\ |y| &= e^{C_1 - C_2}\,e^{-x} \\ y &= A\,e^{-x} \textrm{ where } A = \pm e^{C_1 - C_2} \end{align*}$

There you go, all functions of the form $\displaystyle \begin{align*} f(x) = A\,e^{-x} \end{align*}$ will satisfy $\displaystyle \begin{align*} f(x) + f'(x) = 0 \end{align*}$ .

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