Question

Linear Equations with Constant Coefficients.

Need help please.

Answer 1

Given:
\[(D ^{2}+3D+2)y = 12x^{2}\]

Find the general solution, \[\frac{ dy _{p} }{dx}\], A, B, C and particular solution.
Man this is a lot. x_x

Answer 2

can you find the homogeouse solution y_c ?

Answer 3

so I should solve for the homogenous solution first? o.o

Answer 4

yeah always find the homogenouse solution before the particular solution

Answer 5

\[y_c''+3y_c'+2y=0\\
m^2+3m+2=0\\m_{1,2}=\]

Answer 6

can you do this bit?

Answer 7

\[y_c=Ae^{m_1x}+Be^{m_2x}\]

Answer 8

now for y_p ,

are you going to use ; operator D method,
undertermined coefficients,
or variation of parameters ?

Answer 9

I would like to use operator D method, but.. I need to solve first for the
dy_p / dx.

Answer 10

are you sure that comes first?

how are we to take the derivative of y_p, if we havent found y_p ?

Answer 11

so what should I do? I'm clueless because this was just introduced to us. x_x

Answer 12

have you got y_c yet?

Answer 13

nope. o.o
What should I do first, sir? ._.

Answer 14

start with solving the quadratic

m^2+3m+2=0

Answer 15

oh. okay.

Answer 16

x = -3 +- 1i / 2

Answer 17

try again

Answer 18

okay. :<

Answer 19

oh. sorry. it's m = -1 and -2.

Answer 20

good , so y_c=

Answer 21

what is that y_c? o.o

Answer 22

the complementry homogenouse solution

\[y_c=Ae^{m_1x}+Be^{m_2x}\]

Answer 23

oh. is it the same with C1e^x? o.o

Answer 24

yeah some time the constants are written
A,B,C ,...
or sometimes C_1, C_2, C_3, ...

Answer 25

\[y = C _{1}e ^{-x}+C_{2}e^{-2x}\]

Answer 26

well thats just the complementry homogenouse solution, y_c

y=y_c+y_p


we havent found y_p yet

Answer 27

so y_p = y_c - y? o.o

Answer 28

The general solution is \(y\)

to get the general solution we first find
the complementy homogenouse equation \(y_c\) , like you have,

and then the particular solution \(y_p\). [we haven't found this yet ].

We add these togeter to get the general solution \(y=y_c+y_p\)

Answer 29

how can we find y_p? o.o

Answer 30

Well you said you wanted to use operator D, this is one method to find y_p.

Answer 31

but there's 12x^2 in the way. o.o

Answer 32

if the 12x^2 was a 0, then y_p =0, but its not so we have to find y_p

Answer 33

\[(D^2+3D+2)y_p=12x^2\\
y_p=\frac{1}{D^2+3D+2}12x^2\]

Answer 34

now you are able to factoise that denominator using what did before

\[D^2+3D+2=(D...)(D...)\]

Answer 35

so it's:
12x^2 / (m+2) (m+1) ?

Answer 36

@UnkleRhaukus, sir are you still there?

Answer 37

looks like a tough question, didn't read lol. #yolo

Answer 38

yeah but this time is not m's its D's

Answer 39

\[y_p=\frac1{D+1}\frac1{D+2}12x^2\]

Answer 40

now you have to know the amazing solution to this useful fraction\[\frac1{1-x}=1+x+x^2+x^3+\cdots\]

Answer 41

first look the inner most bit

\[\frac1{D+2}\]
filp it around abit

\[\frac1{2+D}\]

now get rid two out of the fraction

\[\frac12\frac1{1+\frac D2}\]


now try to use the \(\frac1{1-x}=1+x+x^2+x^3+\cdots\) formula,

what will you subsitute for \(x\) ?

Answer 42

2 ? o.o

Answer 43

compare\[\frac1{1+\frac D2}\qquad\qquad \&\qquad\qquad \frac1{1-x}\]

Answer 44

...

Answer 45

I can't understand that part. o.o

Answer 46

well if x=-D/2

you can change \(\frac1{1-D/2}\) into \( 1-D/2+(D/2)^2-(D/2)^3+(D/2)^4-...\)

Answer 47

So,

\[y_p=\frac1{D+1}\frac1{D+2}12x^2\\\qquad\vdots\\\qquad=\frac12\frac1{D+1}(1-D/2+D^2/4-D^3/8+D^4/16-\cdots)12x^3\]

Answer 48

now just look at this bit

\[(1−D/2+D^2/4−D^3/8+D^4/16−⋯)12x^3\]

Answer 49

do you know how to applyu operator D here?

Answer 50

yeah. wait. I'll try to do it. o.o

Answer 51

I mean yes. o.o

Answer 52

it might make it eaisier to find Dx^3, D^2x^3, D^3x^3, D^4x^3
first

(remember that notation just means the deriviatives of x^3 with respect to x

Answer 53

\[\mathrm Dx^3=(x^3)'=\frac{\mathrm d }{\mathrm dx}(x^3)=3\times x^{3-1}=\]

Answer 54

is it really D^3x^3? not D3x^3? o.o

Answer 55

\[\mathrm D^3x^3=(x^3)'''=\frac{\mathrm d^3 }{\mathrm dx^3}(x^3)=\mathrm D^2(3\times x^{3-1})=\]

Answer 56

oh.. okay.

Answer 57

so the answer to the 3rd derivative of x^3 is 6.

Answer 58

yes

Answer 59

what about D^4(x^3) ?

Answer 60

It is 0.

Answer 61

right !

Answer 62

so it's just until D^3.. o.o

Answer 63

what's next after knowing this? o.o

Answer 64

yeah, all the higher order derivativies are zero

Answer 65

so now you can simplify this bit by distrubution

\[(1−D/2+D^2/4−D^3/8+D^4/16−⋯)12x^3=\]

Answer 66

err, I got -3x^2+6x-5.

Answer 67

are you missing the first term ?

Answer 68

12x^3-3x^2+6x-6.

Answer 69

what should we do now to simplify 1/D+1? o.o

Answer 70

ops i messed up i changed 12x^2 to 12x^3 by mistake a while back

Answer 71

oh. okay.
so it's 12x^2 - 2x + 2?

Answer 72

oh wait.

Answer 73

am I right with that 12x^2 - 2x + 2?

Answer 74

hmmm thats not what im getting,

Answer 75

12x^2 - 18x + 6?

Answer 76

thats closer

Answer 77

12x^2 - 12x + 6.

Answer 78

that's what i am getting

Answer 79

its kinda tricky

Answer 80

ok now you have

(1−D/2+D2/4−D3/8+D4/16−⋯)12x2 =12x^2 - 12x + 6.


\[y_p=\frac1{D+1}\frac1{D+2}12x^2\\
\qquad\vdots\\
\qquad=\frac12\frac1{D+1}(1-D/2+D^2/4-D^3/8+D^4/16-\cdots)12x^2\\
\qquad\qquad\vdots\\
\qquad\qquad=\frac12\frac1{D+1}\Big(
12x^2 - 12x + 6\Big)\]

Answer 81

so now it's.. uhh..
\[(1 - D + D ^{2} - D ^{3} + D ^{4} - ... )12x ^{2}-12x +6\] ?

Answer 82

or may just D. o.o

Answer 83

yes, , it migte make it eaisier if you cancle that factor first

Answer 84

i suspose we could have done that ages ago,

Answer 85

so it could've been become
\[\frac{ 1 }{ 2 }(D+1)(D+2)(12x^{2})\]
???

Answer 86

I feel so ashamed of myself. ._.

Answer 87

i mean

\[y_p=\frac1{D+1}\frac1{D+2}12x^2\\
\qquad\vdots\\
\qquad=\frac12\frac1{D+1}(1-D/2+D^2/4-D^3/8+D^4/16-\cdots)12x^2\\
\qquad=\frac1{D+1}(1-D/2+D^2/4-D^3/8+D^4/16-\cdots)6x^2\\
\qquad\qquad\vdots\\
\qquad\quad=\frac1{D+1}\Big(
6x^2 - 6x + 3\Big)\\
\\
\qquad\qquad\vdots\\
\qquad\qquad=\]

Answer 88

so you have

\[(1−D+D^2−D^3+D^4−...)\big (6x^2−6x+3\big )\]

Answer 89

I got 6x^2 - 18x + 21

Answer 90

very good work

Answer 91

so now, putting it all together. . .

This expression we found using operator D and that magic fraction
is your particular solution \(y_p\),

remember we found the complementry homogenouse solution
\(y_c\) earlier

__ __ __
Can you now, tell me the General solution \(y=y_c+y_p\)

to the origional
differential equation \[(D^2+3D+2)y=12x^2\]?

Answer 92

\[y = C _{1}e^{-x} + C_{2}e^{-2x} + 6x^{2} - 18x + 21\]

Answer 93

That is it!

Answer 94

do you kinda understand the procedure now?

Answer 95

that was tough. but yes, I understand it now.

Answer 96

is it always like this when the general solution is asked?

Answer 97

well if you are using operator D, [like we did] and the instead of x^3 being on the right hand side say it is cos x or e^x or some thing , then we wont use the fraction expansion we have to uses something else

Answer 98

it's even more complex, right?