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

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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

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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#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.

Please restate the question, or clear this up.

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#3

well...

you know the auxiliary equation...

so... the CF...

then it follows that y = Ae^px e^qi + Be^px e^-qi

y = e^px (Ae^qi + Be^-qi)

now... from the

we get...

now as

so y = e^px (A(cosqx+isinqx) + B(cosqx-isinqx))

y= e^px (Acosqx + Aisinqx + Bcosqx - Bisinqx)

Then...

Let

so...

QED... btw.. Euler's identity is very hard to prove... so take it as quoted

**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) = -sinqxso 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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#4

(Original post by

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.

**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.

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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#5

(Original post by

btw.. Euler's identity is very hard to prove... so take it as quoted

**anchemis**)btw.. Euler's identity is very hard to prove... so take it as quoted

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.

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#6

(Original post by

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...

**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...

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.

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#7

(Original post by

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.

**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.

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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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#9

(Original post by

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.

**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.

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#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.

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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