Results 1 to 3 of 3

Thread: Ivp

  1. #1
    Junior Member
    Joined
    Sep 2007
    Posts
    36

    Ivp

    Given the IVP $\displaystyle y'' - 4y = 0, y(0) = 2, y'(0) = \alpha$

    1.) Find sol'n of the IVP in terms of $\displaystyle \alpha$

    2.) Find $\displaystyle \alpha$ such that as $\displaystyle t \rightarrow \infty$ the sol'n approached zero.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,002
    Thanks
    1
    First, we find the auxiliary equation by the quadratic.

    $\displaystyle m^{2}-4=(x+2)(x-2)=0$

    This gives us:

    $\displaystyle C_{1}e^{2x}+C_{2}e^{-2x}$

    Now, by using y(0)=2, we can find C1:

    $\displaystyle C_{1}e^{2(0)}+C_{2}e^{-2(0)}=2$

    $\displaystyle C_{1}=2-C_{2}$

    But $\displaystyle y'=2C_{1}e^{2x}-2C_{2}e^{-2x}$

    Then using the other IC, we can find C2:

    $\displaystyle 2(2-C_{2})e^{2(0)}-2C_{2}e^{-2(0)}=a$

    $\displaystyle C_{2}=1-\frac{1}{4}a$

    Then $\displaystyle C_{1}=\frac{1}{4}a+1$

    Therefore, the IVP in terms of a is:

    $\displaystyle (1+\frac{1}{4}a)e^{2x}+(1-\frac{1}{4}a)e^{-2x}$

    Now, find the limit for the last portion of the problem. It should be kind of easy to see what a is now.

    $\displaystyle \lim_{x\rightarrow{\infty}}[(1+\frac{a}{4})e^{2x}]+\lim_{x\rightarrow{\infty}}[(1-\frac{a}{4})e^{-2x}]$

    What is the limit on the right when x is unbounded?. What will give 0?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2007
    Posts
    36
    Regarding the limit... won't this only be 0 when alpha (or a in your case) = -4? That'll take care of the first part and the 2nd part
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum