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Math Help - derivatives

  1. #1
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    derivatives

    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
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  2. #2
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    Re: derivatives

    \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}
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  3. #3
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    Re: derivatives

    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
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  4. #4
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    Re: derivatives

    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))
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  5. #5
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    Re: derivatives

    Quote Originally Posted by thealivision View Post
    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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