Question

Find particular solution \[\large y''+y'+y=14+8x+11\cos(x)\]

Answer 1

I have tackled similar questions before, but this one is a bit of a challenge to me and was hoping for a in-depth insight :)

Answer 2

the first step is to find the corresponding homogenous solution with the auxiliary equation

Answer 3

\[m^2+m+1=0\]

Answer 4

@Mimi_x3 variation of parameters?

Answer 5

yes I had \[\large m^{2}+m+1=0\] as my characteristic equation @UnkleRhaukus as equation (1)
I had \[\large 14+8x+11\cos(x)\] as equation (2)
then I get lost

Answer 6

work out the two values of m using the quadratic formula

Answer 7

that comes out as \[\frac{-1\pm \sqrt3 i}{2}\]

Answer 8

yeah

Answer 9

you have complex solutions for m , so m=a±ib

so the corresponding homogenous solution is

\[y_c=e^{ax}\Big(A\cos(bx)+B\sin(bx)\Big)\]

Answer 10

What do you get for the homogenous solution @Ray10 ?

Answer 11

ooh okay, I haven't used this one before

Answer 12

so \[\large m = \frac{-1 \pm \sqrt3 i}{2}\]

Answer 13

yes so what is a, b?

Answer 14

a would be \[\large \frac{-1+\sqrt3 i}{2}\] ?? Or I am confused if not :S

Answer 15

um no, break the m up into a real term ± and imaginary term

Answer 16

oh now I get it!

Answer 17

a= \[\large \frac{-1}{2}\] and b= \[\large \frac { \pm \sqrt3 i}{2}\]

Answer 18

yeah thats pretty right, but the form was m=a±ib so the b term dosn't have the i in it

Answer 19

so now just sub in a=-1/2 and b=√3/2 into

\[y_c=e^{ax}\Big(A\cos(bx)+B\sin(bx)\Big)\]

Answer 20

if it doesn't have an i in it does that make it: \[\large m= \frac{-1 \pm \sqrt3}{2}, \frac{-1}{2} \pm \frac{ \pm \sqrt3}{2}\] ? in the form of m=a \[a\pm ib\]

Answer 21

like where does the minus of the \[\pm\] go?

Answer 22

well you had m right, m does have an i in it,

\[m=a\pm i b\]


once we have this form we extract the real and plus/minus imaginary parts

the formula for the complementary homogenous solution, only needs a, and b

Answer 23

\[y_{c}=e^{\frac{-1}{2}x}(Acos(\frac{\sqrt3}{2}x))+Bsin(\frac{\sqrt3}{2}x))\] ? :)

Answer 24

ah now that makes sense :)

Answer 25

YES! that's it

Answer 26

:D

Answer 27

So you have the complementary homogenous solution

Answer 28

yes so now that is \[\large h(x)\] the homogenous solution correct?

Answer 29

The particular solution will solve

\[y_p′′+y_p′+y_p=14+8x+11\cos(x)\]

now, it is a good idea to break the particular solution into two parts
\[y_p=y_{p_1}+y_{p_2}\]
\[y_{p_1}''+y_{p_1}'+y_{p_1}=14+8x\]and\[y_{p_2}''+y_{p_2}'+y_{p_2}=11\cos x\]


Now i would recommend undetermined coefficients to find \(y_{p_1}\) and\( y_{p_2}\)

Answer 30

yeah i think we are using different letters for notation

i use \[y(x)=y_c(x)+y_p(x)\]
to say the general solution is the sum of the complementary homogenous, and particular solutions

but you might have y(x)=h(x)+p(x) or somesuch

Answer 31

so
what would be the 'guess' particular solution (with undetermined coefficients)
\[y_{p_1}=h_1(x)=?\]

to
\[y_{p_1}''+y_{p_1}'+y_{p_1}=14+8x\]

Answer 32

oh of course! yes fair enough
so I see how you are looking at it now :)
yes I have y(x)=h(x)+p(x)

Answer 33

ops ** typo

\[y_{p_1}=p_1(x)=?\]

Answer 34

ah I was confused there haha I was trying to work it out as well :P

Answer 35

the particular solution guess then?

Answer 36

Yeah ,

Answer 37

is it \[\large axe^{kx}\] ?
I was looking at the trial solution table

Answer 38

this question is really confusing to me :S

Answer 39

the non-homogenous term in the first particular solution is 14+8x

Answer 40

:O that solution is missing from my table!! :O

Answer 41

and the square box is \[\Large x^{m}\]

Answer 42

its the easiest case, you can probably remember it , it just says if the non-homogenous term is a polynomial, the guess will be a polynomial of the same order,

say the non-homogenous term was 2x^3, the guess would be Ax^3+Bx^2+Cx+D

Answer 43

oh because the power was 3?

Answer 44

yeah,

Answer 45

and if it was \[\large 2x^{4}\] it woud be \[\large Ax^{4}+Bx^{3}+Cx^{2}+Dx+E\]?

Answer 46

that's right,

Answer 47

so what do you get for 14+8x ?

Answer 48

hmmm well its only to the power of one here

Answer 49

correct me if I'm wrong, but does it seem to work in aspect of differentiating?

Answer 50

like following term is m-1

Answer 51

so 14+8x becomes 8?

Answer 52

there will be constants terms in the 'guess' , we have to solve for these in a moment

Answer 53

that is where I get confused with this question
because originally we were working with, getting to \[\large Ae^{kx}+Be^{kx}\]

Answer 54

?

Answer 55

like what will the guess be? how do we work out the guess for this 14+8x

Answer 56

our non homogenoues termi is a first order polynomial,
So the 'guess' or trial solution well be a first order polynomial with two constants

(dont use A, B thought because we already have them in the homogenous solution,
use C, D)

Answer 57

14C+8xD, is what I understand from that, have I gone about it wrong?
should it be C+xD?

Answer 58

yes
our trial solution for the first particular solution will be
\[y_p(x)=p(x)=C+Dx\]

Answer 59

Woooo@@ I'm finally understanding this more :D

Answer 60

Now take the first and second derivative of p(x)

Answer 61

\[x'= D\]\[x''=0\]

Answer 62

yeah but they are p' and p''

Answer 63

ohh of course!
\[\large p'=D\] \[\large p'=0\]

Answer 64

actually p'_1, p_''1,
*

Answer 65

ok so now
sub them into
\[ p_1''+p_1'+p_1=14+8x\]

Answer 66

you mean: \[\large p_{1} ' ,p_{1}''\]?
oh you already are a step ahead :D

Answer 67

\[\large p_{1} \] in this case is 14+8x?

Answer 68

\[\large 0+D+14+8x=14+8x\]?

Answer 69

p_1 was our trial soution

p_1=C+Dx

Answer 70

my mistake

Answer 71

\[\large 0+D+C+Dx=14+8x\]

Answer 72

yes that's better


can you solve for C and D?

Answer 73

I get a general solution answer :S

Answer 74

I'll show you

Answer 75

There is a special technique here you can use ,

its called equating coefficients

\[(C+D)+(D)x=(14)+(8)x\]

Answer 76

oh alright, thank you for showing me that method :)

Answer 77

if follow that

(C+D)=14, D=8

Answer 78

do I get D=8 and C=6?

Answer 79

yes great!

Answer 80

oh this is great!! :D

Answer 81

so our trial solution was
\[p_1=C+Dx\]

we now have C and D

so what is the first particular solution equal to?

Answer 82

\[\large p_{1}=C+Dx=6+8x\]

Answer 83

excellent

now all we need is the second particular solution,
the one that solves ;
\[p_2''+p_2'+p_2=11\cos x\]

Answer 84

What will our trial solution be this time?

Answer 85

thank you :)
Alright so now we must find a particular solution guess correct?

Answer 86

yeah, you can use a table if you need

Answer 87

\[\large c_{1}\cos(ax+b)+c_{2}\sin(ax+b)\] ?

Answer 88

that is the right bit of the table to look at, but,

our non homogenous term was 11cosx

so
\[ c_{1}\cos(x)+c_{2}\sin(x)\]or\[ E\cos(x)+F\sin(x)\]
is fine

Answer 89

because there is no addition in the cos brackets of course :) yes I understand that :)

Answer 90

yeah if you really wanted to you could have those extra constants but it would be tonnes more work and the would turn out to be a=1, b=0

Answer 91

I just differentiated them, is that correct?

Answer 92

Yes, take the first and second derivative of this second particular trial solution ,


then sub them into \[p_2''+p_2'+p_2=11\cos x\],


solve for E, F by equating coefficients ,

Answer 93

for \[\large p_{1}'=E(Fcos(x)-Esin(x))\]\[\large p_{1}''=-E(Ecos(x)+Fsin(x))\]

Answer 94

wait, what?

Answer 95

then it is \[\large p_{1}''=-E(Ecos(x)+Fsin(x))\] +\[\large p_{1}'=E(Fcos(x)-Esin(x))\]+\[\large p_{1}=E(cos(x)+Fsin(x)\]