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Math Help - Second Order Homogeneous Equation IVP

  1. #1
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    Post Second Order Homogeneous Equation IVP

    I am stuck on this question, I have done a little but I am generally lost. The second order equation is the general form of modeling springs (g is for gamma, but g is easier to type)

    ******
    Show that the solution of the initial value problem

    my'' + gy' + ky = 0, y(t0) = y0, y'(t0) = y1

    can be expressed as the sum y = v + w, where v satisfies the initial conditions

    v(t0) = y0, v'(t0) = 0

    w satisfies the initial conditions

    w(t0) = 0, w'(t0) = y1

    and both v and w satisfy the same differential equation as u.
    ******

    What I have done is solved the auxiliary equation mλ + gλ + λ = 0

    Giving, λ = (g/2m)(-1 (1 - (4km/g))^0.5)

    So there are three possible cases:

    1: If g - 4mk > 0

    y = A*exp(λ1 t) + B*exp(λ2 t)

    2: If g - 4mk = 0

    y = (A + Bt)*exp(-gt/2m)

    3: if g - 4mk < 0

    y = exp(-g/2m)*(Acos(vt) + Bsin(vt))

    I really don't know if i'm on the right track, any help would be much appreciated!
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  2. #2
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    Quote Originally Posted by purakanui View Post
    I am stuck on this question, I have done a little but I am generally lost. The second order equation is the general form of modeling springs (g is for gamma, but g is easier to type)

    ******
    Show that the solution of the initial value problem

    my'' + gy' + ky = 0, y(t0) = y0, y'(t0) = y1

    can be expressed as the sum y = v + w, where v satisfies the initial conditions

    v(t0) = y0, v'(t0) = 0

    w satisfies the initial conditions

    w(t0) = 0, w'(t0) = y1

    and both v and w satisfy the same differential equation as u.
    ******
    I'm not the expert in the field but nevertheless I can share my ideas with you and see where that gets you.

    First put to use the fact that both v and w satisfy the same differential equation as y:
    mv'' + gv' + kv = 0 \Rightarrow v''=-\frac{1}{m}(gv'+kv)
    mw'' + gw' + kw = 0 \Rightarrow w''=-\frac{1}{m}(gw'+kw)

    Now look at the dif. eq. in terms of y
    my'' + gy' + ky = 0. If y=v+w is the solution to the equation than using such y should lead to a mathematical identity. First notice that y'=v'+w' and y''=v'' + w'' and use that in the equation:
    m(v''+w'')+g(v'+w')+k(v+w)=0. Now use the expressions for v'' and w''.

    m(-\frac{1}{m}(gv'+kv)-\frac{1}{m}(gw'+kw))+g(v'+w')+k(v+w)=0..

    Everything on the left side cancels out and you have the identity 0=0 which is of course true, thus proving that the solution indeed can be defined as y=v+w with v and w satisfying the same diff. eq. as y.

    Not sure, but maybe it will help you somehow.

    Anyway, v and w here are solutions to the diff. eq. and their sum is also a solution.
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