Help! 2nd order linear DEs!

Question

anonymous · Answer

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accessdenied · Answer

Do you know how to solve the characteristic equation?  The situation from substituting $ y = e^{\lambda x} $ and solving the quadratic in $\lambda$.

Knowing what result comes from different types of solutions would be useful here.

kainui · Answer

Just curious, how do they expect you to figure these out? By drawing the phase plane as a system of linear equations?

accessdenied · Answer

The way I see it, you always have the solution of a form  y=  e^(kx).  The value of k can be found by just substituting it and its derivatives into the equation.  The e^(kx) part is nonzero, so you can divide it off.  That leaves you with the characteristic equation, which basically gives the two solutions of k for that equation and general solution:   y = C1 e^(k1 x). + C2 e^(k2 x).

If k is positive in one or both solutions, the limit as x approaches infinity just makes y explode (even if one is negative, the positive exponent will take over easily)
If k is negative for both, then both tend towards 0 as x approaches infinity (e.g.  e^(-x))
If k is complex/imaginary, then we have a trig function which is periodic.

anonymous · Answer

does that mean i just need to apply the general solutions for 3 types of characteristic eqns? for eg. y"-10y'+21y=0. it has roots of 3 and 7, so its not periodic means it not bounded?

accessdenied · Answer

yeah.   It is not periodic, and also both are positive exponents on the exponential function.
  y = C1 e^(3x) + C2 e^(7x)    <--- as x approaches infinity, both e^(3x) and e^(7x) go to infinity.

if you had -3 and 7,
  y = C1 e^(-3x) + C2 e^(7x)   <-- e^(-3x) goes to 0, but e^(7x) goes off to infinity, so overall it is not bounded.

if you got -3 and -7, to illustrate the other way around,
  y = C1 e^(-3x) + C2 e^(-7x)   <---  this just goes very close to 0 as x gets much bigger.  that should also be 'bounded' by the definition.

anonymous · Answer

ooo, so that means i just have to find roots that are both negative as these solutions will be bounded, is it?

accessdenied · Answer

yep.  also the roots that are complex / imaginary, that covers your periodic functions.  Both negative, or both are imaginary/complex should be bounded.  :)

anonymous · Answer

ohhh thank you soo much @AccessDenied! that is really helpful! :D

accessdenied · Answer

You're welcome!  :D