Question

\[y^{\prime\prime}+4y^\prime=8x+6\sin2x\]

Answer 1

\[y_c^{\prime\prime}+4y_c^\prime=0\]\[\Delta =(4)^2-4(1)=12\]\[m^2+4m=0\]\[m=\frac{-4\pm2\sqrt3}{2}\]\[m_{1,2}=-2\pm\sqrt3\]\[y_c=Ae^{-2+\sqrt3x}+Be^{-2-\sqrt3x}\]

Answer 2

\[y_{p_1}^{\prime\prime}+4y_{p_1}^\prime=8x\]\[y_{p_1}=Cx+D;\qquad y^\prime_{p_1}=C:\qquad y^\prime_{p_1}=0\]\[0+4C=8x;\qquad C=2\]\[y_{p_1}=2x\]

Answer 3

\[y_{p_2}^{\prime\prime}+4y_{p_2}^\prime=6\sin2x\]
\[y_{p_2}=E\sin (Fx)+G\cos(Hx)\]\[y_{p_2}^\prime=EF\cos(Fx) -GH\sin(Hx)\]\[ y_{p_2}^{\prime\prime}=-EF^2\sin(Fx)-GH^2\cos(Hx)\]\[\left(-EF^2\sin(Fx)-GH^2\cos(Hx)\right)+4\left(E\sin (Fx)+G\cos(Hx)\right)=6\sin2x\]\[(4-F^2)E\sin(Fx)+(4-H^2)G\cos(Hx)=6\sin2x\]

Answer 4

now what?

Answer 5

should i have \[F=H\]?

Answer 6

you messed up at the beginning dude

Answer 7

y''+4y'=0
so
r^2+4r=0
the the characteristic polly for the complimentary

Answer 8

i see
i should have\[\Delta=4^2\]

Answer 9

I don't know what that delta notation is
I see though that you do have m^2+4m=0 but you found the solution to be some crazy numbers

Answer 10

i was using delta for the i discriminant

Answer 11

\[m^2+4m=0\implies m(m+4)=0; m=\{0,-4\}\]

Answer 12

that is right

Answer 13

but you wrote something else above with radicals and stuff...
anyway, so the complimentary is\[y_c=c_1+c_2e^{-4x}\]

Answer 14

so for the guess for the particular wee need to add the guesses for the polynomial part and the trig part

Answer 15

the guess for the polynomial part is\[Ax+B\]and the guess for the trig part is\[C\sin(2x)+D\cos(2x)\](you have to keep the 2x, it never changes with derivatives or integrals, right?)

Answer 16

ah good point

Answer 17

so our guess for the particular will be\[Y_p=Ax+B+C\sin(2x)+D\cos(2x)\]

Answer 18

oh wait, but I am missing something
we only have y'' and y' so I think we need to add an extra x...

Answer 19

\[\Delta =(4)^2=16\]\[m^2+4m=0\]\[m(m-4)=0\]\[m_{1,2}=0,4\]\[y_c=A+Be^{4x}\]
how did you get negative in the index

Answer 20

how did you get from\[m^2+4m=0\]to\[m(m-4)=0\]o-0 ?

Answer 21

ah yes

Answer 22

I would suggest, instead of guessing the particular solution, try the annihilator method as given in these notes. They come in pretty handy to solve this sort of problems
http://home.iitk.ac.in/~sghorai/TEACHING/MTH203/ode9.pdf

Answer 23

so yeah, I missed that because we need to keep the solutions linearly independent, and we already have a constant as an answer in the particular, we need to multiply the polynomial part of our guess by x
so it should be\[Y_p=Ax^2+Bx+C\sin(2x)+D\cos(2x)\]

Answer 24

\[y_{p_2}^{\prime\prime}+4y_{p_2}^\prime=6\sin2x\]\[y_{p_2}=E\sin (2x)+G\cos(2x)\]\[y_{p_2}^\prime=2E\cos(2x) -2G\sin(2x)\]\[ y_{p_2}^{\prime\prime}=-4E\sin(2x)-4G\cos(2x)\]

Answer 25

ok...

Answer 26

is annihilator method the same as using operator D?

Answer 27

yes

Answer 28

I am self-taught in DE's, so I don't know that terminology :(

Answer 29

that is amazing turing test

Answer 30

I concur with Unkle :O. I certainly wouldnt have been able to self teach DE's. I'm still not familiar with PDE's

Answer 31

thanks, I'm reading this annihilator thing now
I've seen these D's but never understood what they mean

Answer 32

\[\left(-4E\sin(2x)-4G\cos(2x)\right)+4\left(E\sin (2x)+G\cos(2x)\right)=6\sin2x\]
\[-3G\cos(Hx)=6\sin2x\]

Answer 33

how did that H get in there ....H=2

Answer 34

\[G=-2\tan (2x)\]
have i made more mistakes yet?

Answer 35

In this particular problem, it would seem to make a lot a sense to simply integrate
both sides before doing anything(since the RHS is a derivative.). Then you can apply
your favorite method to the resulting first order eq.....

Answer 36

why diden't i see that earlier @fwizbang .

Answer 37

this isn't going to work
look Unk, you did mess up, everything cancels, even the cos

Answer 38

hmm

Answer 39

-4+4≠3?

Answer 40

oh wait, you put y for y'

Answer 41

shoud be\[\left(-4E\sin(2x)-4G\cos(2x)\right)+4\left(2E\cos (2x)+2G\sin(2x)\right)=6\sin2x\]

Answer 42

-sin I mean..

Answer 43

i did too, how embarrassing, failure to substitute correctly on my part

Answer 44

\[\left(-4E\sin(2x)-4G\cos(2x)\right)+4\left(2E\cos (2x)-2G\sin(2x)\right)=6\sin2x\]

Answer 45

\[\left(-4E\sin(2x)-4G\cos(2x)\right)+4\left(2E\cos(2x) -2G\sin(2x)\right)=6\sin(2x)\]

Answer 46

good they match ,

Answer 47

Short and sweet using Annihilator's .. Basically \(D \equiv d/dx\)
For the particular solution, \(y_{p}\), it must satisfy
\((D^{2}+4D)y_{p} = 8x+6sin2x\)
Now to annihilate \(8x\), we operate \(D^{2}\) and to annihilate \(\sin{2x}\) we operate \(D^{2}+4\), so operating \(D^{2}(D^{2}+4) \) with the DE, we get

\( D^{3}(D+4)(D^{2}+4)y_{p} = 0\)
this gives \(y_{p} = c_{1} + c_{2}x + c_{3}x^{2} + c_{4}e^{-4x}+c_{5}\sin{2x}+c_{6}\cos{2x}\)

Since, \(c_{1} + c_{4}e^{-4x} \) form part of the homogenous solution, therefore the particular solution \(y_{p} = c_{2}x + c_{3}x^{2}+c_{5}\sin{2x}+c_{6}\cos{2x} \)

Answer 48

ill try annihilation with operator D after i know i can finish with equating co-efficient

Answer 49

I don't get the D^2+4 to annihilate 6sin(2x) could someone explain that?

Answer 50

oh wait I think I get it...

Answer 51

\[\frac{d^2}{dx^2}(6\sin2x)+4(6\sin(2x)=0\]is that the idea?

Answer 52

cool

Answer 53

The annihilator in essence is related to the solution of the characteristic equations.
Note that sin(2x) will occur as a solution only if the roots of the characteristic equation are \(\pm 2i\). i.e. the characteristic equation is \(m^{2}+4\), consequently meaning that if we operate \{D^{2}+4\) on \(\sin{2x}\) we get 0.

Answer 54

yes @TuringTest

Answer 55

gotchya, thanks :) I get it now

Answer 56

\[\left(-4E\sin(2x)-4G\cos(2x)\right)+4\left(2E\cos(2x) -2G\sin(2x)\right)=6\sin(2x)\]
\[-4(E+G)\sin(2x)+4(2E-G)\cos(2x)=6\sin(2x)\]
have i got this step right?

Answer 57

ah no i see an other mistake already

Answer 58

the first G in the second line should have a 2 in front of it,

Answer 59

\[-4(E+2G)\sin(2x)+4(2E-G)\cos(2x)=6\sin(2x)\]
what should my next move be

Answer 60

oh, set the coefficients equal\[-4(E+2G)=6\]\[4(2E-G)=0\]and solve the system

Answer 61

so from the your second equation G=2E

Answer 62

I guess, yeah
do you see where I got these equations from?

Answer 63

yeah because these are the coefficients of the trig functions , and must be equal

Answer 64

yeah

Answer 65

\[G=-\frac{3}{10}\]?

Answer 66

erm... either G or E is -3/10
I'm getting a little confused :S

Answer 67

yeah I get E=-3/10

Answer 68

ok,

Answer 69

you were right i did get E/G mixed up

Answer 70

\[E=-\frac{3}{10};\qquad G=-\frac 35\]

Answer 71

yep, so now we move on to finding the values of D and E (or whatever you called them) for the other particular

Answer 72

oh you may want to write out this particular in full just so you don't forget which coefficient goes where

Answer 73

\[y_{p_2}=-\frac 3{10}\sin (2x)-\frac5{10}\cos(2x)\]

Answer 74

yep, so now we do \(y_{p1}\)

Answer 75

i think i have \( y_{p_1} \) in my second post for this question

Answer 76

oh but it is wrong

Answer 77

it's wrong because the solution you guessed is not linearly independent from the complimentary, as I mentioned earlier

Answer 78

hmm i have made so many mistakes

Answer 79

we already know the complimentary is\[y_c=Ae^{-4x}+B\]so if\[y_p=Cx+D\]then we have two constants, B and D, hence the solutions are not linearly independent
to fix this problem we need to multiply our guess for \(y_p\) by x

Answer 80

\[y_{p1}=Cx^2+Dx\]now our complimentary and particular solutions are linearly independent, as they should be

Answer 81

"if you don't make mistakes it means you didn't try hard enough" is a paraphrase of a favorite quote of mine

Answer 82

so we are multiplying by x because the original equation is \(y\)-absent?

Answer 83

\[y_{p_1}=Cx^2+Dx;\qquad y^\prime_{p_1}=2Cx+D:\qquad y^{\prime\prime}_{p_1}=2C\]
\[2C+4Cx^2+Dx=8x;\qquad D=\frac 12\]

Answer 84

we need the solutions to be linearly independent
in other words we have to keep the wronskian from being zero
if we allow one of our solutions to be linearly independent on the other (the constant solution in the complimentary, A is linearly dependent on part of our guess for the particular, Cx+D) so this would not be a fundamental set of solutions
multiplying by x fixes that problem

Answer 85

you could say that it is "because we don't have a y" in this particular problem, but a better way to think of it is that our guess for the particular cannot have any linear dependence on the complimentary solution, so when we see
"oh, I have a constant in my complimentary and a constant for my guess for the particular, I need to fix that"
no part of the complimentary should be a constant multiple of the particular

Answer 86

and D should be -1/2

Answer 87

you forgot to multiply D by the 4

Answer 88

hahah

Answer 89

actually, a lot of that is wrong; I think you are being lazy, distracted or tired...
I've seen you do much harder stuff\[y''+4y'=2C+8Cx+4D=8x\]

Answer 90

yeah its 417am, i am not tired enough to be asleep

Answer 91

that would explain it
look over this when you have your wits about you
http://tutorial.math.lamar.edu/Classes/DE/UndeterminedCoefficients.aspx

Answer 92

turns out im not very at substitution

Answer 93

lol

Answer 94

\[y_{p_1}^{\prime\prime}+4y_{p_1}^\prime=8x\]
\[y_{p_1}=Cx^2+Dx;\qquad y^\prime_{p_1}=2Cx+D:\qquad y^{\prime\prime}_{p_1}=2C\]
\[2C+8Cx+4D=8x\]\[2(C+2D)=0;\qquad 8Cx=8x\]\[C=1;\qquad D=-\frac 12\]

Answer 95

so
\[y_c=A+Be^{-4x}\]\[y_{p_1}=x^2-\frac x2\]\[y_{p_2}=-\frac{3}{10}\left(\sin 2x+2\cos2x\right)\]
\[y=y_c+y_{p_1}+y_{p_2}\]
\[y(x)=A+Be^{-4x}+x^2 -\frac x2-\frac{3}{10}\left(\sin 2x+2\cos2x\right)\]

Answer 96

all done ,

fwizbang get the prize for simplest solution