Results 1 to 4 of 4

Thread: please help me with this question

  1. #1
    Junior Member
    Joined
    Mar 2009
    Posts
    66

    please help me with this question

    i need help with the following question attached.

    thanks
    Attached Thumbnails Attached Thumbnails please help me with this question-question.jpg  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    6
    The equation is...

    $\displaystyle y^{'}-2\cdot y=-5\cdot x$ (1)

    ... and its caratteristic equation...

    $\displaystyle a-2=0$

    ... has solution $\displaystyle a=2$. So the general integral of 'incomplete equation' is...

    $\displaystyle y_{i}(x)= c\cdot e^{2x}$ (2)

    It is aesy to verify that a particular integral of (1) is...

    $\displaystyle y_{p}(x)= -\frac{5}{2}\cdot x -\frac{5}{4}$ (3)

    ... so that the general integral of (1) is...

    $\displaystyle y(x)= c\cdot e^{2x} -\frac{5}{2}\cdot x -\frac{5}{4}$ (4)

    If we insert the 'initial condition' we obtain...

    $\displaystyle y(x)= \frac{17}{4}\cdot e^{2x} -\frac{5}{2}\cdot x -\frac{5}{4}$ (5)

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    848
    Hello, cooltowns!

    It's fairly straight forward . . . Where is your difficulty?


    Solve: .$\displaystyle \frac{dy}{dx} - 2y \:=\:5x,\quad y(0) = 3$

    Integrating factor: .$\displaystyle I \:=\:e^{\int \text{-}2\,dx} \:=\:e^{\text{-}2x}$

    We have: .$\displaystyle e^{-2x}\frac{dy}{dx} - 2e^{-2x} y \;=\;5xe^{-2x} $

    . . . . . . . $\displaystyle \frac{d}{dx}\left(e^{2x}y\right) \:=\:5xe^{-2x}$

    Integrate: . $\displaystyle e^{2x}y \;=\;5\!\!\int\!\! xe^{-2x}dx $


    . . By parts: .$\displaystyle \begin{array}{ccccccc}u &=& x & & dv &=& e^{-2x}dx \\ \\[-4mm] du &=& dx & & v &=&-\frac{1}{2}e^{-2x} \end{array}$


    We have: .$\displaystyle e^{-2x}y \;=\;5\bigg[\text{-}\tfrac{1}{2}xe^{-2x} + \tfrac{1}{2}\!\!\int e^{-2x}dx\bigg] \;=\;5\bigg[\text{-}\tfrac{1}{2}xe^{-2x} -\tfrac{1}{4}e^{-2x}\bigg] + C $

    . . . . . . . $\displaystyle e^{-2x}y \;=\;-\tfrac{5}{2}xe^{-2x} - \tfrac{5}{4}e^{-2x} + C $


    Multiply by $\displaystyle e^{2x}\!:\quad y \;=\;-\tfrac{5}{2}x - \tfrac{5}{4} + Ce^{2x} $

    Since $\displaystyle y(0)=3\!:\quad3 \;=\;-\tfrac{5}{2}(0) - \tfrac{5}{4} + Ce^0 \quad\Rightarrow\quad C \:=\:\frac{17}{4}$


    Therefore: . $\displaystyle y \;=\;-\frac{5}{2}x - \frac{5}{4} + \frac{17}{4}e^{2x} $

    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Mar 2009
    Posts
    66
    i haven't done this type of differential equation before i didn't know you have use intergrating factors.
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum