# Math Help - Second Order Homogeneous Equation IVP

1. ## 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)

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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.
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What I have done is solved the auxiliary equation mλ² + gλ + λ = 0

Giving, λ = (g/2m)(-1 ± (1 - (4km/g²))^0.5)

So there are three possible cases:

1: If g² - 4mk > 0

y = A*exp(λ1 t) + B*exp(λ2 t)

2: If g² - 4mk = 0

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

3: if g² - 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!

2. Originally Posted by purakanui
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.