Question

y'' + 9y = 2xsinx + xe^(3x)
Need to solve this using method of undetermined coefficients. What I need to know: coefficients of particular solutions. For example, the "A" coefficient for first term (2xsinx) is 1/4. What is B since in my particular solution for 2xsinx (without the second term -- xe^(3x)) is (Ax+B)cosx+(Ax+B)sinx. Please help!

Answer 1

\[ \frac{d^2y}{dx^2} + 9 \frac{dy}{dx} = 2xsinx + xe^{3x}\]Just to make it look better. : )

Answer 2

Ok, thank you. Have any idea for a solution?

Answer 3

I am trying. : )

Answer 4

http://www.wolframalpha.com/input/?i=y%22+%2B9y%27+%3D+2xsinx+%2Be^%283x%29

Out of my League Sorry.

Answer 5

I assume you can solve for fundamental solution and only wants particular soloution?

Answer 6

Yes. I can find the homogenous solution which is : A*cos(3x) + B*sin(3x)

Answer 7

\[My particular solution: (AX+B)cosx+(AX+B)sinx. After differentiating the particular solution and putting the particular solution along with its derivatives \in the main equation (but with only 2xsinx) I get: 8AXsinx+8Acosx-2Asinx+8Bsinx+8Bcosx+2Acosx=2xsinx. From this I get A=1/4. What is b?\]

Answer 8

Myparticularsolution:(AX+B)cosx+(AX+B)sinx.After differentiating the particular solution and putting the particular solution along with its derivatives the main equation (but with only 2xsinx) I get: 8AXsinx+8Acosx−2Asinx+8Bsinx+8Bcosx+2Acosx=2xsinx. From this I get A=1/4.What is B?

* 2 minutes ago

Answer 9

You can also just try to solve: \[y \prime \prime + 9y = 2xsinx\]
All I need is the coefficient B of the particular solution which is:
\[(AX+B)cosx + (AX+B)sinx\]
If somebody knows this please help.

Answer 10

give me a sec i'm looking at it k?

Answer 11

ok

Answer 12

give me a second and i will check my work

Answer 13

ok.

Answer 14

oops i forgot about that 9

Answer 15

you know the one in front of the y
i forgot about that one
that changes things

Answer 16

Does the x in xsin x mess things up at all?

Answer 17

in front of sin x..

Answer 18

give me a few more minutes to redo it

Answer 19

@myininaya Shouldn't you have a x term before cosx, like \[(x/4)cosx\]?

Answer 20

Maybe (dunno) u have to multiply the trial sol'n by x...

Answer 21

@estudier. No it doesn't. You make a particular solution to be: (AX+B)cosx + (AX+B)sinx, then differentiate, then put that in the original equation and you should get A and B from it.

Answer 22

Yes, that might work, I suppose...

I'm only used to doing the pieces separately.

Answer 23

How u doing, myinanaya?

Answer 24

okay
\[y_p=-\frac{1}{16}\cos(x)+\frac{1}{4}xsin(x)+\frac{-1}{54}e^{3x}+\frac{1}{18}xe^{3x}\]

Answer 25

now let me check this lol

Answer 26

:-)

Answer 27

Look: \[8AXsinX+8AcosX-2AsinX+8BsinX+8BcosX+2AcosX=2XsinX\]
Now, from this since there is only one term on the left that corresponds to \[XsinX\] \[8A = 2\]\[A = 1/4\] Since there is no terms on the right with \[cosX\] or \[sinX\] we set sum of every coefficient on the left that has corresponding sinX or cosX to 0 (zero). What is B? Is it 0 (zero) or is it something elese?

Answer 28

@myininay: Could be. Are you sure in the answer? So So you got A= -(1/16) and B = 1/4. 1/4 is what I got here too but 1/16 the missing coefficient I couldn't solve.

Answer 29

i'm checking it
it takes alittle time for me to check sorry

Answer 30

That's fine. Take your time.

Answer 31

for undetermined coefficients, you hypothesize a particular solution that is a linear combination of the RHS plus all its derivatives. For example,
2xsinx --> Axsinx+Bcosx (if I did that right)
xe^3x --> Cxe^3x + De^3x
plug all of that into the differential eq, and match coeff to the RHS

Answer 32

2xsinx --> Axsinx+Bcosx

I was wondering whether it had to be xCosx as well.

Answer 33

@phi I know. That's why I asked people to solve only with 2xsinx on the right. What I want to know is how to handle coefficient B. I know that for general particular slution I need to add both particular soultions.

Answer 34

yes thats right
i got it right :)

Answer 35

Hold on! The way I laearned is that for a 2XsinX the particular solution is: (AX+B)cosX + (AX+B)sinX.

Answer 36

myininaya Your particular solution for 2XsinX is like mine: (AX+B)cosX + (AX+B)sinX?

Answer 37

The way I learned was if A CosX + B sin X wasn't going to work, then multiply it by X (possibly more than once9.

Answer 38

:)

Answer 39

@estudier That is if term in particular guess is similar to the homogeneous solution.

Answer 40

check those pdfs aarnes

Answer 41

the first one shows how i got the particular solution and the 2nd pdf shows me checking

Answer 42

it was tedious
every move matters
that is why it took me so long

Answer 43

OK, here's just yp= Axsinx + Bcosx

Answer 44

y'= Axcosx + Asinx - Bsinx

Answer 45

@ I'm checking them right now myininaya.

Answer 46

y''= Acosx - Axsinx + Acosx-Bcosx

Answer 47

plug those into y'' + 9y= 2xsinx
you should get A= 1/4 and B= -1/16

Answer 48

and phi is probably right you don't need that whole general particular solution we used but as long as you do things correctly it doesn't matter as long as we have the parts that do matter

Answer 49

I guess if u do these all the time u learn where the shortcuts are....

Answer 50

yay phi! :)

Answer 51

we got the same thing
but i took a longer route

Answer 52

Educational.....

Answer 53

thanks...

Answer 54

So is op happy?

Answer 55

op?

Answer 56

original poster

Answer 57

oh stayed tuned...

Answer 58

Whit! @myinimaya: Is this solution to homgeneous eq: http://www.mediafire.com/i/?4ek64ieay6t3l2m

Answer 59

whit?

Answer 60

Wait. :)

Answer 61

Yep. That's what wolfram says...

Answer 62

looks good

Answer 63

Here's how I would do the exponential part:

the derivatives of y= \( x e^{3x} \) are
\[ y'=3xe^{3x} + e^{3x}\]
\[y''=3xe^{3x} + 3e^{3x}+3e^{3x}\]
if we keep going, we will see that all the derivatives will be, neglecting constant coefficients, of the form
\[ C xe^{3x} + De^{3x} \]
C and D to be determined.
Assume \( yp=C xe^{3x} + De^{3x} \)

plug yp into the differential equation, group like terms, and solve for C and D.
If you want I'll grind through it...

Answer 64

Maybe i's late but here's what a I got: http://www.mediafire.com/i/?juqxn6ro5979oex
I'm guessing it's wrong since my only coefficients for first particular solution are A and B. I didn't have C and D. The reason I didn't had them is because of this website:
http://tutorial.math.lamar.edu/Classes/DE/UndeterminedCoefficients.aspx#Second_UnderCoeff_Ex6b
Your thoughts?

Answer 65

Your yp2 solution looks good.

Your starting yp1 does not look right. Each term must have its own coefficient.
Instead of (Ax + B)(A cos(x) + B sin(x))
you should use
yp1= A x cos(x) + B x sin(x) + C cos(x) + D sin(x)

Answer 66

To find an assumed solution, find all the derivates of the RHS= 2 x sin(x)
We can ignore the constant 2 (we will be using undetermined coeffs)
x sin(x)
x cos(x) + sin(x)
-x sin(x) + cos(x) + cos(x)
it should be clear that if we continue taking derivatives, we will have only terms
sin(x), cos(x), x sin(x), x cos(x) neglecting any constant coefficients.

So assume yp1= A x cos(x) + B x sin(x) + C cos(x) + D sin(x)


Let's see how it turns out.
Take the derivative of our assumed solution yp1:
yp1= A x cos(x) + B x sin(x) + C cos(x) + D sin(x)
yp1' = -A x sin(x) + A cos(x) +B x cos(x) + B sin(x) -C sin(x) + D cos(x)
yp1''= -A x cos(x) - A sin(x) - A sin(x) - B x sin(x) + B cos(x) + B cos(x) - C cos(x) - D sin(x)

y'' +9y = 2x sin(x)
(-A+9A) x cos(x) +(-B+9B) x sin(x) + (2B -C+9C) cos(x) + (-2A-D+9D) sin(x)= 2x sin(x)

8A=0, A=0
8B= 2, B= 1/4
2B+8C=0, C= -1/16
8D=0, D=0

so yp= (1/4) x sin(x) - (1/16) cos(x)

Answer 67

As a check, we can use wolfram
http://www.wolframalpha.com/input/?i=solve+y%27%27+%2B+9y+%3D+2xsinx

Answer 68

So assume yp1= A x cos(x) + B x sin(x) + C cos(x) + D sin(x)

The key part, same as op but with constants absorbed.

Answer 69

aarnes started with (Ax +B)(A cos(x) + Bsin(x))
see http://www.mediafire.com/i/?juqxn6ro5979oex

Answer 70

yp1= A x cos(x) + B x sin(x) + C cos(x) + D sin(x)
-> cosx(Ax +C) + Sinx(Bx+D)

Difference is absorbed constants.

Answer 71

So if there is a polynomial times a trig function we use particular solution AX*cosX + BX*sinX + CcosX + D sinX? Just how did we get to here, phi?

Answer 72

I tried to explain it in the above post.
Another way, is starting with your
(Ax +B)(A cos(x) + Bsin(x))
multiply it out to get A^2 x cos(x) +ABcos(x) +ABx sin(x) +B^2sin(x)
and then rename the coeffs. They are all different.

Answer 73

rename the coeffs. They are all different.

Yes, I think that was the cause of the problem (confusing).

Answer 74

Oh I get it now. From (AX+B)(CcosX+DsinX) we get ACXcosX +ADXsinX+BCcosX+BDsinX then : AXcosX+BXsinX + CcosX + DsinX

Answer 75

Yes. That works.

Answer 76

Great. then what does this website talks about: http://tutorial.math.lamar.edu/Classes/DE/UndeterminedCoefficients.aspx#Second_UnderCoeff_Ex6b

Answer 77

Quoting "So, to avoid this we will do the same thing that we did in the previous example. Everywhere we see a product of constants we will rename it and call it a single constant."
He is doing exactly what you just did. Rename the coeffs.

Answer 78

Yes, I think so too, he is usually pretty reliable, that fellow...

Answer 79

Aaah. Yes he did. My mistake then. I was fast reading and didn't see that he introduced D E and F constants. Oh man. :D

Answer 80

Well, we can call this topic closed! :) Thank you all for help.