Proof - Complex 2nd Order Diff Equ......Watch

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Thread starter 14 years ago
#1
Hi,

in maths today we attempted, along with the teacher, to help prove that for any complex complementary function, the sum will always = 0 for any value of the constant multiplying that function, if you catch my drift.

it is said that for a complex number complementary function y = e^(px) . (C . cos(qx) + D . sin(rx)) = 0 This is a standard way of expressing this.

Prove that for the equation:

a . d2y/dx^2 + b . dy/dx + ky = 0

For any values of the constants C and D?

We got so far but ended up with to many unknowns, we were able to prove this for y = e^(px)cos(qx) but not for the above. If the theory could be proved it would be much appricated!

Many Thanks

Streety
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14 years ago
#2
I'm not entirely sure what you are saying here. You seem to be saying in the above equation that y = 0. How can e^(px) . (C . cos(qx) + D . sin(rx)) = 0 ?

Please restate the question, or clear this up.
0
14 years ago
#3
well... a d2y/dx2 + b dy/dx + cy = f(x)

you know the auxiliary equation... ak^2+bk+c = 0 --> for complex k = p + qi

so... the CF... y = Ae^(p+qi)x + Be^(p-qi)x
then it follows that y = Ae^px e^qi + Be^px e^-qi
y = e^px (Ae^qi + Be^-qi)

now... from the Euler's Identity... e^ix = cosx + isinx

we get... y = e^px (A(cosqx+isinqx) + B(cos(-qx)+isin(-qx)))

now as cos is even... cos(-qx) = cosqx
sin is odd.. so sin(-qx) = -sinqx

so y = e^px (A(cosqx+isinqx) + B(cosqx-isinqx))
y= e^px (Acosqx + Aisinqx + Bcosqx - Bisinqx)

Then... y = e^px ((A+B)cosqx + (A-B)i sinqx)
Let A+B = C and (A-B)i = D

so... y = e^px (Ccosqx + Dsinqx)

QED... btw.. Euler's identity is very hard to prove... so take it as quoted
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14 years ago
#4
(Original post by AntiMagicMan)
I'm not entirely sure what you are saying here. You seem to be saying in the above equation that y = 0. How can e^(px) . (C . cos(qx) + D . sin(rx)) = 0 ?

Please restate the question, or clear this up.
err... second order differential equations consist of a CF and a PI...

you find CF by solving the Auxiliary Equation... the whole LHS = 0
and if roots are distinct k1, k2 --> y = Ae^k1x + Be^k2x
if roots are repeated k --> y = (Ax+B)e^kx
if roots are complex as k = p+qi --> y = e^px (Ccosqx+Dsinqx)

then you find the PI according to the form of the Auxiliary equation... try out values for PI and general solution is CF+PI...
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14 years ago
#5
(Original post by anchemis)
btw.. Euler's identity is very hard to prove... so take it as quoted
It's not hard at all...

Recall that the series expansions of the following functions are as follows:

exp(x) = 1 + x + (x^2)/2! + (x^3)/3! + (x^4)/4! + ...
sin(x) = x - (x^3)/3! + (x^5)/5! - (x^7)/7! + ...
cos(x) = 1 - (x^2)/2! + (x^4)/4! - (x^6)/6! + ...

So,

exp(ix) = 1 + ix + ([ix]^2)/2! + ([ix]^3)/3! + ([ix]^4)/4! + ...

=> 1 + ix - (x^2)/2! - i(x^3)/3! + (x^4)/4! + ...
=> [1 - (x^2)/2! + (x^4)/4! - ...] + [ix - i(x^3)/3! + ...]
=> [1 - (x^2)/2! + (x^4)/4! - ...] + i[x - (x^3)/3! + ...]
=> cos(x) + isin(x)

QED

It's just a matter of knowing the series expansions.

Galois.
0
14 years ago
#6
(Original post by anchemis)
err... second order differential equations consist of a CF and a PI...

you find CF by solving the Auxiliary Equation... the whole LHS = 0
and if roots are distinct k1, k2 --> y = Ae^k1x + Be^k2x
if roots are repeated k --> y = (Ax+B)e^kx
if roots are complex as k = p+qi --> y = e^px (Ccosqx+Dsinqx)

then you find the PI according to the form of the Auxiliary equation... try out values for PI and general solution is CF+PI...
Oh wow, you can solve second order differential equations. Hooray.

Anyway, back to my original question, which is what does "for any complex complementary function, the sum will always = 0 for any value of the constant multiplying that function". Is the question "is the sum of the constants C and D always zero"? That question at least makes sense, whereas the question being asked of us currently makes no sense to me as it stands.
0
14 years ago
#7
(Original post by AntiMagicMan)
Oh wow, you can solve second order differential equations. Hooray.

Anyway, back to my original question, which is what does "for any complex complementary function, the sum will always = 0 for any value of the constant multiplying that function". Is the question "is the sum of the constants C and D always zero"? That question at least makes sense, whereas the question being asked of us currently makes no sense to me as it stands.
yep no idea what he's on about....
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Thread starter 14 years ago
#8
What i mean is that for any C.F. that involves complex functions, prove that like any other function (expressing it using expontials, cos & sine) that when used back into the original equation it will always = 0 regardless of the values of the constant multipuls C & D.
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14 years ago
#9
(Original post by streetyfatb)
What i mean is that for any C.F. that involves complex functions, prove that like any other function (expressing it using expontials, cos & sine) that when used back into the original equation it will always = 0 regardless of the values of the constant multipuls C & D.
You want to prove if y = e^(rx)[Acos(px) + Bsin(qx)], then a(d²y/dx²) + b(dy/dx) + ky = 0?
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14 years ago
#10
Look like you want to make things more complicated.
0
Thread starter 14 years ago
#11
thats what i want to prove DVS
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14 years ago
#12
Hmm.. I don't think that's the case, generally speaking. It is however true if and only if b²<ac when ay''+by'+cy=0. To show this solve the equation.

ay''+by'+cy=0

Auxilary equation:
am²+bx²+c=0
m=[-b +/- sqrt(b²-4ac)]/2a

Since we assumed that b²<4ac, then let sqrt(b²-4ac)=di. That transforms the roots into:
m=(-b +/- idi)/2a=p +/- qi.

So the complementary function is:
y=e^(px) [Acos(qx)+Bsin(qx)], which is the form required.
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