Results 1 to 5 of 5

Thread: I lied, one more (first-order DE). :)

  1. #1
    Junior Member cinder's Avatar
    Joined
    Feb 2006
    Posts
    60

    I lied, one more (first-order DE). :)

    $\displaystyle x\frac{dy}{dx}+y=\sin(x)$, $\displaystyle x>0$, initial condition: $\displaystyle y(\frac{\pi}{2})=1$

    What I have so far...

    $\displaystyle \frac{dy}{dx}+\frac{1}{x}y=\frac{\sin(x)}{x}
    \Rightarrow \int \frac{1}{x}dx=\ln|x| \Rightarrow e^{\ln|x|}=x
    $

    I'm not sure if I did the previous part right. And after that I'm unsure how to proceed.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    5
    Quote Originally Posted by cinder View Post
    $\displaystyle x\frac{dy}{dx}+y=\sin(x)$, $\displaystyle x>0$, initial condition: $\displaystyle y(\frac{\pi}{2})=1$

    What I have so far...

    $\displaystyle \frac{dy}{dx}+\frac{1}{x}y=\frac{\sin(x)}{x}
    \Rightarrow \int \frac{1}{x}dx=\ln|x| \Rightarrow e^{\ln|x|}=x
    $

    I'm not sure if I did the previous part right. And after that I'm unsure how to proceed.
    $\displaystyle x\frac{dy}{dx}+y=\sin(x)$

    may be written as:

    $\displaystyle \frac{d}{dx} (x~y) =\sin(x)$

    so:

    $\displaystyle x~y = -\cos(x) +C$

    and the rest should be simple.

    RonL
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,002
    Thanks
    1
    You're OK so far. You have your integrating factor.

    $\displaystyle \frac{d}{dx}(xy)=sin(x)$

    Integrate:

    $\displaystyle \int\frac{d}{dx}(xy)=\int{sin(x)}dx$

    $\displaystyle xy=-cos(x)+C$

    $\displaystyle y=\frac{-cos(x)}{x}+\frac{C}{x}$

    IC is $\displaystyle y(\frac{\pi}{2})=1$

    Using this and solving for $\displaystyle C=\frac{\pi}{2}$

    Therefore, you have:

    $\displaystyle y=\frac{\pi}{2x}-\frac{cos(x)}{x}$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member cinder's Avatar
    Joined
    Feb 2006
    Posts
    60
    I guess I'm not seeing how it can be rewritten as $\displaystyle \frac{d}{dx}(xy)=sin(x)$.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    11,163
    Thanks
    736
    Awards
    1
    Quote Originally Posted by cinder View Post
    I guess I'm not seeing how it can be rewritten as $\displaystyle \frac{d}{dx}(xy)=sin(x)$.
    Think of it this way:
    $\displaystyle \frac{d}{dx}(xy) = 1 \cdot y + x \frac{dy}{dx}$
    by the product rule.

    So
    $\displaystyle \frac{d}{dx}(xy) = y + x \frac{dy}{dx} = sin(x)$
    which is your initial equation.

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Re-writing higher order spatial derivatives as lower order system
    Posted in the Differential Equations Forum
    Replies: 11
    Last Post: Jul 27th 2010, 08:56 AM
  2. Replies: 1
    Last Post: Oct 27th 2009, 04:03 AM
  3. Replies: 2
    Last Post: Feb 23rd 2009, 05:54 AM
  4. Replies: 2
    Last Post: Jan 9th 2009, 07:31 PM
  5. Replies: 2
    Last Post: Nov 25th 2008, 09:29 PM

Search Tags


/mathhelpforum @mathhelpforum