Question

y''+y'+y = 2+x+cos x
The question is find a particular integral for it, how do I begin?

Answer 1

1/ find general solution
2/ find particular
do 1 first.

Answer 2

okey so the general solution is a*y'' +b*y' +c*y, a,b and c are constants and the sum of the particular integral and the complementary function is y = y(pi)+ y(cf) and y(cf) contains no arbitrary constants and y(ct) contains two arbitrary constants, how do go about?

Answer 3

Characteristic equation for homogeneous part is r^2+r+1 =0 gives us roots r =\(-\frac{1}{2}\pm i\sqrt3\) --> the general solution is???

Answer 4

I'm totally lost

Answer 5

You have to do part 1 to find general solution because it tells you what particular solution is. That's why I ask you do that.

Answer 6

okey so y = eâx(Acos Bx +Bsin Bx)?

Answer 7

so that gives me e^-.5x(Acos 1,23x + B sin 2,23x)?

Answer 8

In this case, the general solution has the form of \(y = e^{-x/2}(A cos (\sqrt3 x)+B sin(\sqrt 3 x)\)

Answer 9

yup

Answer 10

Now, non-homogenous part: what is the form of the solution??

Answer 11

don't forget you have 2 parts: polynomial and trig
for polynomial, do as usual
for trig must times x to make it different from general solution

Answer 12

okey Ill try to see if I can solve it now, hopyfully I'll get it right.

Answer 13

Okey if y = e−x/2(Acos(3√x)+Bsin(3√x), and y = y(pi) + y(cf)
What is y(pi) ?? @OOOPS

Answer 14

it says y(pi) = 1+x+sinx, but how??

Answer 15

I don't know your notations. But I guess y (pi) is y (general) and y(cf) is y (particular) right?

Answer 16

yeah

Answer 17

ok, for non-homogenous part, you have 2 parts : polynomial + trig
1/ for polynomial, let's call the solution of this part is y1 , find it out by solving
y" +y'+y = 1+x
the solution y1= ax^2+bx+c
y1' =
y"=
--------------------------
you work on it

Answer 18

Know how to do??

Answer 19

okey so y1' is = 2ax+b
but y''= that s the tricky part

Answer 20

y"= 2a , why is it hard?

Answer 21

did u mean y''=y1''?

Answer 22

oh in that case it wasnt hard, I thought for something else okey, yeah but now im with you.

Answer 23

no, hihi
since y1 is solution, right? so that y1 satisfies the equation
y1" + y1'+y1=1+x

Answer 24

and from above, we replace all
ax + 2ax+b+ax^2+bx+c= 1+x
One more question: do you have initial values?? like y(0) = .... y(1) =...

Answer 25

nope i dont.

Answer 26

it just says find the particular integral.

Answer 27

Wow!! so everything is on letter form?? I mean ax +b......

Answer 28

yeah it goes like this
Q = y''+y'+y = 2+x+cos x
A = y(particular) = 1+ x+ sin x

Answer 29

ok, back to the problem, got me so far from the ax+2ax +b+ax^2 +bx +c =1+x

Answer 30

now how do we make that into 1 + x + sin x? (Im starting to feel so stupid)

Answer 31

We MUST break it into 2 parts; I just solve the first part: polynomial

Answer 32

from the above equation: we have
a=0
3a+b=1 --> b =1
b+c=1 --> c =0
got it??

Answer 33

ah yeah I think so, and where does the sin x come in?

Answer 34

so, y1= ax^2+bx+c= x
the first partial solution is x
now the second part
solve y2 = ecosx + fsinx
y2'=?
y2"= ?
----------------------------
??

Answer 35

okey. y2' = cos x but why fsinx tho?
y2'' ? -sin x

Answer 36

I use e, f as coefficients to differentiate from coefficients above. We used A, B, a, b,c already, I don't want to use them again. That's it

Answer 37

straight up, this is so hard, because our test is in another language, so the terms and everything will be more confusing, we do everthihng in english and have exams in swedish...

Answer 38

The standard form of partial solution for sin x, cos x, is Asinx + B cosx
solve from it

Answer 39

ah okey, then it goes -esin x + fcos x
y'' is -ecos x -fsin x

Answer 40

hihihi

Answer 41

this may seem far fetched , I was an A student until I stopped caring...man its hard getting back to being so good..

Answer 42

y2= esinx + fcosx
y2'= ecosx-fsinx
y2"= -esinx -fcosx
------------------------
y2"+y2'+y2= ecosx -fsinx= cosx
which gives us e =1 f =0, --> so y2= cosx
COMBINE y(partial) = y1+y2= x +cos x
done

Answer 43

hey, friend. I am vietnamese. When I study this stuff in America, I struggled a lot with my English and Math in English. hihihiihi

Answer 44

thanks for the looong patience. maybe I'm tired or something, I'll take a break and hopefully get back on track. Thanks alot for all the help and patience :)

Answer 45

ok.