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complementary function

x'' + 2x' + 3x = sin (wt)
Let xc=ceλxx_c = ce^{\lambda x}, so
λ2+2λ+3=0\lambda^2 + 2\lambda + 3 = 0
Therefore λ1=4i1,λ2=4i+1\lambda_1 = 4i - 1, \lambda_2 = 4i + 1

Could anyone possibly tell me what the complementary function then is? My instinct is to use xc=c1cos(wt)+c2sin(wt)x_c = c_1cos (wt) + c_2sin (wt) but I'm not sure when it's imaginary numbers whether you have to sub these in somewhere?? thanks in advance!

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Reply 1
If m = p +/- iq

complementary function

y = e^{px}(Acos qx + Bsin qx)
Reply 2
You've solved the quadratic equation incorrectly.
Reply 3
Heres the Summary to my FP1 differential equations stuff.

1) For the second order differential equation

ad2ydx2+bdydx+cy=0\displaystyle a\frac{d^2y}{dx^2} + b\frac{dy}{dx} + cy = 0

the auxiliary quadratic equation is

am2+bm+c=0\displaystyle am^2 + bm + c = 0

i) If the auxiliary quadratic equation has real distinct roots α\displaystyle \alpha and β \beta (condition b2>4ac\displaystyle b^2 > 4ac ) then the general solution is

y=Aeαx+Beβx\displaystyle y = Ae^{\alpha x} + Be^{\beta x}

Where A and B are constants

ii) If the auxiliary quadratic equation has real coincident roots α \alpha (condition b2=4ac\displaystyle b^2 = 4ac ) then the general solution is

Unparseable latex formula:

\displaystyley = (A+Bx)e^{\alpha x}



Where A and B are constants

iii) If the auxiliary quadratic equation has pure imaginary roots ±ni \pm ni arising from the m^2 + n^2 = 0. The general solution is

y=Acosnx+Bsinnx\displaystyle y = Acos nx + Bsin nx

Where A and B are constants and n is of the element of real numbers.

iv) If the auxiliary quadratic equation has complex conjugate roots p±iq p \pm iq where p,q are of the real set of numbes. (condition b2<4ac\displaystyle b^2 < 4ac) the general solution is

y=epx(Acosqx+Bsinqx)\displaystyle y = e^{px} ( Acos qx + Bsin qx )

Where A and B are constants

2) For the differential equation

ad2ydx2+bdydx+cy=f(x)\displaystyle a\frac{d^2y}{dx^2} + b \frac{dy}{dx} + cy = f(x)

Where A,B,C are constants. The complementary function is the general solution of the differential equation ad2ydx2+bdydx+cy=0\displaystyle a\frac{d^2y}{dx^2} + b\frac{dy}{dx} + cy = 0 and a particular integral is any solution (ie the function of x ) that satisfies the differential equation

ad2ydx2+bdydx+cy=f(x)\displaystyle a\frac{d^2y}{dx^2} + b \frac{dy}{dx} + cy = f(x)

The general solution of the differential equation is

y = complementary function + particular integral.

f(x) could be of the forms

i) a constant k
ii) a linear function px + q
iii) an exponential function Kepx Ke^{px}
iv) a trigonometric function eg psinx , qcos2x or psin3x + pcos3x
v) a quadratic function p + qx + rx^2

For example
d2ydx2+3dydx+2y=f(x)\displaystyle \frac{d^2y}{dx^2} + 3\frac{dy}{dx} + 2y = f(x)

Find the particular integral of the differential equations in the case where f(x) =
a) 12
b) 3x + 5
c) 3e^{2x}
d) cos2x
e) 5 + 12x + 2x^2

a) Try y = k as the particular integral

Unparseable latex formula:

\displaystyle \frac{dy}{dx} = 0 \hspace{10} \frac{d^2y}{dx^2} = 0



Substituting back in

0+0+2k=12 0 + 0 + 2k = 12

k=6 k = 6

So y = 6 is the particular integral

b) Try y = ax + b as the particular integral

Unparseable latex formula:

\displaystyle \frac{dy}{dx} = a \hspace{10} \frac{d^2y}{dx^2} = 0



0+3a+2(ax+b)=3x+5\displaystyle 0 + 3a + 2(ax+b) = 3x + 5

Equating x coefficients
Unparseable latex formula:

\displaystyle 2a = 3 \hspace{10} a = \frac{3}{2}


Equating constant terms
Unparseable latex formula:

\displaystyle 3a + 2b = 5 \hspace{10} \frac{9}{2} + 2b = 5 \hspace{10} b = \frac{1}{4}



so y=32x+14\displaystyle y = \frac{3}{2}x + \frac{1}{4}

c) try y=ke2x\displaystyle y = ke^{2x} as the particular integral

Unparseable latex formula:

\displaystyle \frac{dy}{dx} = 2ke^{2x} \hspace{10} \frac{d^2y}{dx^2} = 4ke^{2x}



Substitute back into the differential equation

4ke2x+6ke2x+2ke2x=3e2x\displaystyle 4ke^{2x} + 6ke^{2x} + 2ke^{2x} = 3e^{2x}

12k=3\displaystyle 12k = 3

k=14\displaystyle k = \frac{1}{4}

The particular integral is y=14e2x\displaystyle y = \frac{1}{4}e^{2x}

d) Try y = acos 2x + b sin 2x

dydx=2asin2x+2bcos2x\displaystyle \frac{dy}{dx} = -2asin2x + 2b cos2x

d2ydx2=4acos2x4bsin2x\displaystyle \frac{d^2y}{dx^2} = -4acos2x - 4b sin2x

Substituting back into the differential equation

4acos2x4bsin2x+3(2asin2x+2bcos2x)+2(acos2x+bsin2x)=cos2x\displaystyle -4acos2x - 4bsin2x + 3(-2asin2x + 2bcos2x) + 2(acos2x + bsin2x) = cos2x

Equating terms in cos2x: 4a+6b+2a=1\displaystyle -4a + 6b + 2a = 1
Equating terms in sin2x: 4b6a+2b=0\displaystyle -4b - 6a + 2b = 0

Solving equations (1) and (2) simultaneously gives

Unparseable latex formula:

\displaystyle a = -\frac{1}{2} \hspace{10} b = \frac{3}{20}



The particular integral is

y=120cos2x+320sin2x\displaystyle y =-\frac{1}{20}cos2x + \frac{3}{20} sin 2x

e) Try y=p+qx+rx2\displaystyle y = p + qx + rx^2

dydx=q+2rx\displaystyle \frac{dy}{dx} = q + 2rx

d2ydx2=2r\displaystyle \frac{d^2y}{dx^2} = 2r

Substituting back into the equation

2r+3(q+2rx)+2(p+qx+rx62)=5+12x+2x2\displaystyle 2r + 3(q+2rx) + 2(p + qx + rx62) = 5 + 12x + 2x^2

That is: 2r+3q+2p+(6r+2q)x+2rx2=5+12x+2x2\displaystyle 2r + 3q + 2p + (6r+2q)x + 2rx^2 = 5 + 12x + 2x^2

Equating x2 x^2 coefficients:
Unparseable latex formula:

\displaystyle 2r = 2 \hspace{10} r = \frac{1}{2}


Equating x coefficients: 6r+2q=12\displaystyle 6r + 2q = 12
r = 1 so
Unparseable latex formula:

\displaystyle 6 + 2q = 12 \hspace{10} q = 3



Equating constant coefficients 2r+3q+2p=5\displaystyle 2r + 3q + 2p = 5

But r = 1 and q = 3 so

2+9+2p=5\displaystyle 2 + 9 + 2p = 5

p=3\displaystyle p = -3

so the particular integral is

y=3+3x+x2\displaystyle y = -3 + 3x + x^2

I hope that helps
Reply 4
Someone's bored...
Reply 5
Speleo
Someone's bored...


Someone should be doing something else other than maths :p:.
Reply 6
I would have loved it if your browser crashed or you hit back by accident towards the end of that post...
:p:
Reply 7
Rabite
I would have loved it if your browser crashed or you hit back by accident towards the end of that post...
:p:


Now that wouldnt be very nice would it! I did hit backspace a few times which made it go back a page. Fortunately for me firefox saved the work i did on this page.
Reply 8
Also, if the particular integral is already a solution to the homogeneous equation (i.e. ad2ydx2+bdydx+cy=0a\frac{d^2 y}{dx^2} + b\frac{dy}{dx} + cy = 0), then you multiply by a power of x. E.g. if y=Aexy = Ae^x is what you get for your particular integral, but it's a solution to the homogeneous equation, then you use y=Axexy = A x e^x, etc.
Reply 9
Thanks SO MUCH insparato you don't know how much of a help that is! But I'm still a little stuck now on what to use for the particular integral...
I've tried xp = Acoswt + Bsinwt
and xp = t{Acoswt + Bsinwt}

but both times didn't really work out, and I couldn't get it into the form of (3w)2sin(wt)2wcos(wt)(3w2)2+4w2\frac{(3-w)^2 sin(wt) - 2wcos(wt)}{(3-w^2)^2 + 4w^2} like the question asks. So I'm just checking to see whether I'm just doing it wrong with the right particular integral, or if everything is just WRONG WRONG!
Reply 10
I don't know how you get those values for your solutions to the auxiliary quadratic; I get λ=2±i82\lambda = \frac{-2 \pm i\sqrt{8}}{2}.
Reply 11
Yep I realised that before :redface: but that doesnt affect the particular integral bit does it?
You have to understand that i am an A level student. I do not know very much about differential equations. So i dont understand the meaning of homogenous and inhomogenous differential equations. So im sorry about the briefness of the stuff i provided. However i hope that helps solve some of your particular integral problems.
hohoho11
Thanks SO MUCH insparato you don't know how much of a help that is! But I'm still a little stuck now on what to use for the particular integral...
I've tried xp = Acoswt + Bsinwt
and xp = t{Acoswt + Bsinwt}

but both times didn't really work out, and I couldn't get it into the form of (3w)2sin(wt)2wcos(wt)(3w2)2+4w2\frac{(3-w)^2 sin(wt) - 2wcos(wt)}{(3-w^2)^2 + 4w^2} like the question asks. So I'm just checking to see whether I'm just doing it wrong with the right particular integral, or if everything is just WRONG WRONG!
A cos wt + B sin wt works. It's a bit of an algebra grind, so just try to be careful and methodical and not miss any factors of w.
insparato
You have to understand that i am an A level student. I do not know very much about differential equations. So i dont understand the meaning of homogenous and inhomogenous differential equations. So im sorry about the briefness of the stuff i provided. However i hope that helps solve some of your particular integral problems.

Aaron, as I have learnt it a homogenous DE is a one where you can sub f(tx,ty) instead of f(x,y) and it makes no difference, like dydx=xyx\frac{dy}{dx}=\frac{x-y}{x}
Reply 15
darth_vader05
I think in this case homogenous means that the RHS equals zero.


Yes.
Reply 16
OK I have got to (3w)2cos(wt)2wsin(wt)w3+12w212w\frac{(3-w)^2 cos(wt) - 2wsin(wt)}{-w^3 + 12w^2 - 12w} and this seems sort of close if you swap over the cos and sin wt and ignore the denominator :cool: but I can't see where I have gone wrong, or how to manipulate this further to get it to what the questions asks...so I'm kind of in limbo...thanks again to everyone who has helped me with this.
We put x=Acosωt+Bsinωtx=A \cos \omega t + B \sin \omega t.

Then x=Aωsinωt+Bωcosωtx' = -A \omega \sin \omega t + B \omega \cos \omega t

And x=Aω2cosωtBω2sinωtx'' = -A \omega^2 \cos \omega t - B \omega^2 \sin \omega t

So x+2x+3x=[(3ω2)A+2ωB]cosωt+[(3ω2)B2ωA]sinωtx''+2x'+3x = [(3-\omega^2)A + 2 \omega B]\cos \omega t + [(3-\omega^2)B - 2 \omega A] \sin \omega t.

Since we want x+2x+3x=sinωtx''+2x'+3x = \sin \omega t we must have:

(3ω2)A+2ωB=0,(3ω2)B2ωA=1(3-\omega^2)A+2\omega B = 0, \quad (3-\omega^2)B-2\omega A = 1

So (3ω2)2B2ω(3ω2)A=3ω2(3-\omega^2)^2B-2\omega(3-\omega^2) A = 3-\omega^2 and (3ω2)A=2ωB(3-\omega^2)A=-2\omega B.

So (3ω2)2B+4ω2B=(3ω2)(3-\omega^2)^2B+4\omega^2B = (3-\omega^2), so B=3ω2(3ω2)2+4ω2B = \frac{3-\omega^2}{(3-\omega^2)^2+4\omega^2}

Once we have B, A is just 2ω3ω2B\frac{-2\omega}{3-\omega^2}B.

Substitute back in x=Acosωt+Bsinωtx=A \cos \omega t + B \sin \omega t to get

x=(3ω2)sinωt2ωcosωt(3ω2)2+4ω2 x=\frac{(3-\omega^2) \sin \omega t -2\omega \cos \omega t}{(3-\omega^2)^2+4\omega^2} as required.
Reply 18
THANK YOU THANK YOU DFranklin, I've got my exam tomorow and thanks to you I'm now feeling very confident!
Thats good to hear :smile:.