Results 1 to 5 of 5

Math Help - Second order linear equations, reduction of order

  1. #1
    Member
    Joined
    May 2008
    Posts
    87

    Second order linear equations, reduction of order

    From Boyce & DiPrima's "Elementary Differential Equations and Boundary Value Problems" book:

    pp. 171, 172. Example 3:

    Given that y1(t) = t^(-1) is a solution of 2t^2y'' + 3ty' - y = 0, t > 0, (31)
    find a second linarly independent solution.

    We set y = v(t)t^(-1); then

    y' = v't^(-1) - vt^(-2), y'' = v''t^(-1) - 2v't^(-2) + 2vt^(-3)

    Substituting for y, y', and y'' in Eq. (31) and collecting terms, we obtain

    2t^2(v''t^(-1) - 2v't^(-2) + 2vt^(-3)) + 3t(v't^(-1) - vt^(-2)) - vt^(-1)
    = 2tv'' + (-4+3)v' + (4t^(-1) - 3t^(-1) - t^(-1))v
    = 2tv'' - v' = 0, (32)

    Note that the coefficient of v is zero, as it should be, this provides a useful check on our algebra.

    Separating the variables in Eq. (32) and solving for v'(t), we find that

    v'(t) = ct^(1/2), (33)

    ...

    I have difficulty following the step from 32 to 33.
    Could someone perform the steps in between 32 and 33 in detail for me so I can see what is actually taking place there?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member flyingsquirrel's Avatar
    Joined
    Apr 2008
    Posts
    802
    Hello,
    Quote Originally Posted by posix_memalign View Post
    = 2tv'' - v' = 0, (32)

    Note that the coefficient of v is zero, as it should be, this provides a useful check on our algebra.

    Separating the variables in Eq. (32) and solving for v'(t), we find that

    v'(t) = ct^(1/2), (33)

    ...

    I have difficulty following the step from 32 to 33.
    Could someone perform the steps in between 32 and 33 in detail for me so I can see what is actually taking place there?
    Let u=v'. (32) can be written 2t\cdot \frac{\mathrm{d}u}{\mathrm{d}t}=u(t).

    Separating the variables : \frac{\mathrm{d}u}{u}=\frac12\cdot \frac{\mathrm{d}t}{t}

    Integrating both sides :

    \ln |u(t)|=\frac{1}{2}\ln t + K=\ln \sqrt{t}+K \implies |u(t)|=\underbrace{\exp K}_{\text{constant}} \sqrt{t}\implies u(t)=v'(t)=C\sqrt{t}

    The last implication comes from that for t>0, \sqrt{t}>0 so u(t) never equals 0 hence u (which is continuous) does not change sign on (0,\infty). Is it clearer ?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    May 2008
    Posts
    87
    Quote Originally Posted by flyingsquirrel View Post

    \ln |u(t)|=\frac{1}{2}\ln t + K=\ln \sqrt{t}+K \implies |u(t)|=\underbrace{\exp K}_{\text{constant}} \sqrt{t}\implies u(t)=v'(t)=C\sqrt{t}

    Is it clearer ?
    Much clearer yes, thanks a lot!

    However, in the following: \ln |u(t)|=\frac{1}{2}\ln t + K=\ln \sqrt{t}+K \implies |u(t)|=\underbrace{\exp K}_{\text{constant}} \sqrt{t}

    There is at first a constant "K" which comes into play as an addition, but after the implication it is a multiplication, why is this?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member flyingsquirrel's Avatar
    Joined
    Apr 2008
    Posts
    802
    Quote Originally Posted by posix_memalign View Post
    However, in the following: \ln |u(t)|=\frac{1}{2}\ln t + K=\ln \sqrt{t}+K \implies |u(t)|=\underbrace{\exp K}_{\text{constant}} \sqrt{t}

    There is at first a constant "K" which comes into play as an addition, but after the implication it is a multiplication, why is this?
    Because \exp(a+b)=\exp a \times \exp b :

    \ln |u(t)|=\ln \sqrt{t}+K \implies \exp\left[\ln |u(t)|\right]=\exp\left[ \ln \sqrt{t}+K\right]

    but \exp\left[ \ln \sqrt{t}+K\right]=\exp\left[\ln \sqrt{t}\right]\exp K =\sqrt{t}\exp K hence we get |u(t)|=\sqrt{t}\exp K .
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by posix_memalign View Post
    Much clearer yes, thanks a lot!

    However, in the following: \ln |u(t)|=\frac{1}{2}\ln t + K=\ln \sqrt{t}+K \implies |u(t)|=\underbrace{\exp K}_{\text{constant}} \sqrt{t}

    There is at first a constant "K" which comes into play as an addition, but after the implication it is a multiplication, why is this?
    e^{\ln \sqrt{t} + K} = e^{\ln \sqrt{t}} e^{K}

    using the usual index law

    = \sqrt{t} e^K
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. 2nd Order, Homog., Reduction of Order
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: November 27th 2011, 07:36 AM
  2. reduction of order with homogeneous equations...
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: February 28th 2011, 04:02 AM
  3. Replies: 1
    Last Post: November 30th 2009, 10:54 AM
  4. A linear first order equations
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 19th 2008, 10:11 PM
  5. Replies: 1
    Last Post: July 29th 2007, 03:37 PM

Search Tags


/mathhelpforum @mathhelpforum