Linearity, fundemental set to DE
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Linearity, fundemental set to DE

[From: ] [author: ] [Date: 11-11-06] [Hit: ]
Im guessing theres an easier way to check for linear independance, I just cant see it yet and have a bit of trouble understanding the question. Particulally:*when they say constant coefficient equation, do they mean something like; Ay + By + Cy + Dy = E .........
Can { exp(2t), exp(t/4), exp(-t)*cos(t), exp(-t)sin(t) } be the fundamental set of solutions to a constant coefficient fourth order linear ODE (ordinary differential equation)?

I tried making a wronskian which gave me a 4 x 4 determinate which im pretty sure is the wrong approach. Im guessing theres an easier way to check for linear independance, I just cant "see" it yet and have a bit of trouble understanding the question. Particulally:

*when they say constant coefficient equation, do they mean something like;
Ay'''' + By''' + Cy'' + Dy' = E ...or
Ay'''' + By''' + Cy'' + Dy' = 0
* and if they asked whether four functions can be the fundamental set to a fifth order differential equation, how does that differ? is there a trick?

Any help would be much appreciated :)

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They are talking about a homogeneous equation, so it has the second form

Ay'''' + By'' + Cy' + Dy = 0.

It is important to know that every nth order linear homogeneous equation, constant coefficient or not, gives rise to a fundamental solution set as long as the coefficients are continuous and the leading coefficient is never zero (on the interval of definition). A fundamental solution set must contain n linearly independent functions that solve the equation.

If the coefficients are real constants, then all solutions will be exponentials, sines and cosines, polynomials or products of these types of functions. Moreover, sines and cosines must occur together.

Your four functions have the right form. You can use the Wronskian to show that they are in fact linearly independent and that W is never zero. So, yes this can be a fundamental solution set for a fourth order constant coefficient equation.

You can actually build the equation if you want to. The characteristic polynomial will have roots
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