Question

how to eliminate the arbitrary constant of this equation y= Ae^(2x) + Bcosx + Cxsinx to find the the d.e.

Answer 1

I think you need to differentiate it 3 times since you have 3 arbitrary constants

Answer 2

@hoho23 please try to calculate \[\frac{ d ^{2}y }{ dx ^{2}}\]

Answer 3

y" = 4Aex^(2x) - B cosx - Cxsinx + 2Ccosx

Answer 4

@eliassaab y" - 4y ?????

Answer 5

@Michele_Laino then what's next??

Answer 6

\[
y''(x)-4 y(x)=-5 \cos (x) (B+C x)-2 C \sin (x)
\]

Answer 7

I'm trying to calculate the other higher derivatives of y

Answer 8

You can choose A, B and C to be anything you want say A=B=C=1 and your DE will be
\[
y''(x)-4 y(x)=-2 \sin (x)-5 (x+1) \cos (x)
\]

Answer 9

@eliassaab can i have a favor?? can you pls. show me the step by step how you come up with that answer?? plsssss :))

Answer 10

i already computed the 5th derivative but i really don't know what to do next T.T

Answer 11

y''''-y=15Ae^(2x)-4Ccosx

Answer 12

and 2y''+2y=10Ae^(2x)+4C cosx

Answer 13

You do not need to compute more than the second derivative.
when you compute
y''- 4 y
you eliminate A
and you can create many D.E that have y as a solution.
As I said A, B and C can be any constants

Answer 14

@hoho23 I can write this:
\[y ^{iv}+2y ^{ii}+y=25Ae ^{2x}\]
may be useful?

Answer 15

For example
\[
y=e^{2 x}+x \cos (x)+\cos (x)
\]
is a solution of
\[
y''-4 y=-5 \cos (x) (1+ x)-2 \sin (x)
\]

Answer 16

So there are infinite number of choices, unless you did not state the problem right.

Answer 17

Does it say what order is the DE

Answer 18

y'(0)=2A

Answer 19

@eliassaab no my professor said that we only need to eliminate the arbitrary content

Answer 20

I thought 3 arbitrary constants require a 3rd order DE ?

Answer 21

constant rather

Answer 22

You have to eliminate them to be a solution of a certain DE. Did he give you the DE?

Answer 23

Otherwise, the problem has infinite number of solutions

Answer 24

i was thinking a second order DE would never require more than two arbitrary constants

Answer 25

You see, when you solve a second order DE, you first solve the DE with the right hand side equal to zero and then you guess a special solution for the given right side.

Answer 26

yeah with a second degree eqn, wouldn't the DE satisfy only a specific set of values for A,B,C from the given general solution ? we can't call them arbitrary anymore right ?

Answer 27

im not so sure i understand the question clearly

Answer 28

*second order eqn

Answer 29

http://www.math.psu.edu/tseng/class/Math251/Notes-2nd%20order%20ODE%20pt2.pdf

Answer 30

how about this one y= Ae^(2x)cos3x + Be^(2x)sin3x

Answer 31

@ganeshie8 pls help me with my follow up question :))

Answer 32

I think your ODE is:
\[y ^{ii}-4y ^{i}+13y=0\]

Answer 33

@hoho23

Answer 34

@Michele_Laino :))) thanks... michele what is your answer to my first question???

Answer 35

I tryed, nevertheless I don't found anything still

Answer 36

@hoho23 for your first question, i'm not sure, nevertheless try this:
\[y ^{iii}-2(1+i)y ^{ii}-(1+4i)y ^{i}+2y=0\]
where i is such that: \[i ^{2}=-1\]
namely the immaginary unit

Answer 37

sorry the third coefficient is -(1-4i) and not -(1+4i)

Answer 38

@Michele_Laino is that 3rd derivative or literally y^iii

Answer 39

it is the third derivative, namely \[\frac{ d ^{3}y }{ dx ^{3} }\]

Answer 40

ah okay :))) thanks

Answer 41

thanks!

Answer 42

@Michele_Laino thanks :)) hope for more help in the future :))

Answer 43

Ok!

Answer 44

@hoho23 Ok!, thanks!

Answer 45

I think the DE has to have real coefficient.