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Math Help - Proving Differential Equation

  1. #1
    Junior Member Coop's Avatar
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    Question Proving Differential Equation

    Hi,

    This is first post, will I be able to post in the normal section after this?
    EDIT: ^Ignore :P I figured it out.

    Anyway, one of my homework problems reads like this

    "Verify that y = -tcos(t)-t is a solution to the initial value problem

    t(dy/dt) = y + sin(t); y(pi)=0
    "

    I solved it,

    dy/dt = (-tcos(t)-t)' = t(sin(t))-cos(t)-1

    After plugging y and y' in,

    t(tsin(t) - cos(t) - 1) = (-tcos(t) - t) + t2sin(t)

    Simplifying,

    t2sin(t) - tcos(t) - t = t2sin(t) - tcos(t) - t

    And,

    -tcos(t) - t = -tcos(t) - t

    But my question is, why did they give me the information that y(pi) = 0? Although it's true, am I supposed to use it anywhere? Is it just extra info. to throw me off?
    Last edited by Coop; February 1st 2013 at 10:41 AM.
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Proving Differential Equation

    The initial value gives you a point through which the solution must pass. I would verify that first:

    y(\pi)=-\pi\cos(\pi)-\pi=\pi-\pi=0

    So, the given solution does satisfy the initial condition.

    Now we need to verify the given solution also satisfies the given ODE:

    y=-t\cos(t)-t=-t(\cos(t)+1)

    Differentiate with respect to t:

    \frac{dy}{dt}=(-t)(-sin(t))+(-1)(\cos(t)+1)=t\sin(t)-\cos(t)-1

    If we multiply through by t, we find:

    t\frac{dy}{dt}=t^2\sin(t)-t\cos(t)-t=t^2\sin(t)+y

    Hence, the given solution is not a solution to the stated IVP, but rather is a solution to

    t\frac{dy}{dt}=y+t^2\sin(t) where y(\pi)=0
    Thanks from Coop and topsquark
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  3. #3
    MHF Contributor

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    Re: Proving Differential Equation

    Quote Originally Posted by Coop View Post
    Hi,

    This is first post, will I be able to post in the normal section after this?
    EDIT: ^Ignore :P I figured it out.

    Anyway, one of my homework problems reads like this

    "Verify that y = -tcos(t)-t is a solution to the initial value problem

    t(dy/dt) = y + sin(t); y(pi)=0
    "

    I solved it,

    dy/dt = (-tcos(t)-t)' = t(sin(t))-cos(t)-1

    After plugging y and y' in,

    t(tsin(t) - cos(t) - 1) = (-tcos(t) - t) + t2sin(t)

    Simplifying,

    t2sin(t) - tcos(t) - t = t2sin(t) - tcos(t) - t

    And,

    -tcos(t) - t = -tcos(t) - t

    But my question is, why did they give me the information that y(pi) = 0? Although it's true, am I supposed to use it anywhere? Is it just extra info. to throw me off?
    They are NOT "giving" you that information. That is part of what you are asked to prove. The problem asks for you to show that y= -t cos(t)- sin(t) satisfies both ty'= y+ sin(t) and y(0)= 0. What you did was prove the first part but you must also show that y(0)= -(0)cos(0)- sin(0)= 0.
    Thanks from Coop and topsquark
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  4. #4
    Junior Member Coop's Avatar
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    Feb 2013
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    New York, USA
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    Re: Proving Differential Equation

    Oh okay thanks guys . And @Mark, my mistake, the equation you gave that y was a solution to was the actual one in the book. I wrote it down wrong here! And @HallsofIvy, thanks for pointing that out, I did not know that. I just stared with DE's last week and my teacher hasn't really gone over proving them yet.
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