Results 1 to 4 of 4

Math Help - 2 Differential Equations

  1. #1
    Junior Member
    Joined
    Jan 2009
    Posts
    36

    2 Differential Equations

    1) Solve the following differential equation
    sin x(e^y + 1) = e^y(1 + cos x) \frac{dy}{dx}
    y(0)=0

    My steps
    -I rearranged the formula and got \frac{sinx}{1+cosx} dx = \frac{e^y}{e^y + 1}dy
    -But Don't know if this is right or how to even take the integral of both sides.

    2)A particle moving along the x-axis encounters a resisting force that results in an acceleration of a=\frac{dv}{dt}=-0.02v^2. Given that x=0 cm and v=35 cm/s at t=0.Find the velocity and the position function for t as a function of t for t is greater than or equal to 0.

    My Steps
    -a=-.02v^2
    -.02\int v^2 dv= -.02 v^3/3
    Is this irhgt. This doesn't seem right.

    Thanks for the help in advance.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Sep 2008
    Posts
    12
    NUMBER 1.) Well I have no clue if you are supposed to set up that diff. eq like that (I don't know any about them). But those are very easily integrated.

    \int{\frac{\sin{x}}{1+\cos{x}}dx} =

    u = 1 + cos(x)
    du = -sin(x)

    so:

    - \int{\frac{-\sin{x}}{1+\cos{x}}dx} = -ln(1 + \cos{x}) + c


    AND -


    \int{\frac{e^y}{e^y + 1}dy} =

    u = e^y + 1
    du = e^y

    so:

    \int{\frac{e^y}{e^y + 1}dy} = ln(e^y + 1) + c


    NUMBER 2.) It seems odd... but appears that it should be correct? The integral of the acceleration function is the velocity function, which is what you did. Integrate that again to get the position function. But it does look odd, especially as the variable inside the function is a v. Shouldn't it be a t?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jan 2009
    Posts
    36
    yea i know that is why i am un easy about doing the problems. they just seem obscure.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Jul 2008
    Posts
    21
    I think you'd rather want to integrate with respect to t than v. Read te problem again.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. 'Differential' in differential equations
    Posted in the Differential Equations Forum
    Replies: 0
    Last Post: October 5th 2010, 10:20 AM
  2. Replies: 2
    Last Post: May 18th 2009, 03:49 AM
  3. differential equations
    Posted in the Calculus Forum
    Replies: 7
    Last Post: May 20th 2008, 07:30 PM
  4. Replies: 5
    Last Post: July 16th 2007, 04:55 AM
  5. Replies: 3
    Last Post: July 9th 2007, 05:30 PM

Search Tags


/mathhelpforum @mathhelpforum