Question

How can I solve this y=C1e^2x cos3x+C2e^3x sin3x?

Answer 1

@ikram002p

Answer 2

\[\Large\rm y=c_1 e^{2x}\cos 3x+c_2 e^{2x}\sin3x\]This is probably the general solution to a second order differential equation.
Do you have initial conditions to find a specific solution?

Answer 3

solve what lol ?
u mean how to find c1, C2
whats the origion equation ?

Answer 4

zepdrix i don't have the initial conditions :(

Answer 5

Then this `is` your solution :o
It's a general solution.
You can't go any further than that without initial conditions D:

Answer 6

@ganeshie8

Answer 7

yeah or type the origion Qn tht include y'' ,y' ,y

Answer 8

ahm thankyou for answering my question zepdrix
:) I have another question can you please help me to solve ? :)

Answer 9

I think she wants to eliminate the constants

Answer 10

ganeshie8 You're right I just want to eliminate the constants

Answer 11

okay... since you have two constants - you need to dfferentiate it twice

Answer 12

ganeshie08 can you please help me to solve that question?

Answer 13

differentiating and solving looks complicated >.<
@ikram002p any shortcut ?

Answer 14

you can turn it into a DE to eliminate the parameter constants

Answer 15

like ganeshie already said xD

Answer 16

That will lead you back to the differential equation that corresponded to this solution. Is that what you're trying to do vanessa?

Answer 17

y=Ae^2x + Bxe^2x How about this ? Can you solve this?

Answer 18

zepdrix I just want to eliminate the constants.

Answer 19

a repeated root lead to this solution

Answer 20

you can work back to see what the differential equation must have been

Answer 21

That equation will not have these constants of variation in it yet

Answer 22

what must your characteristic equatoin have been to get these solutions

Answer 23

This is my first time
studying D.E

Answer 24

okay lets look at a homogenous 2nd order linear equation

Answer 25

dan815 please help me :)

Answer 26

y''+ay'+by=0

Answer 27

Thanks dan815 :)

Answer 28

we say that y=e^rx is a solution and try to find what r has to be

Answer 29

y'=re^rx
y''=r^2e^rx
y''+ay'+by=0
e^rx(r^2+ar+b)=0
now this is what your solution was
y=Ae^2x + Bxe^2x
this happens when you have a reapeated root of r=2

Answer 30

so r^2+ar+b = (r-2)^2

Answer 31

anr (r-2)^2=r^2-4r+4
therefore a= -4 and b=4

Answer 32

your equation must have been y''-4y'+4y=0
This was a possible Differential equation

Answer 33

which no longer has those variation parameters

Answer 34

I just want to eliminate the constants

Answer 35

we did, this is the only way I can think of without having BVP or IVPs

Answer 36

Would you please give me a simple way to solve that?

Answer 37

hmm I think they want you to notice that the equation must have been from a repeated root that was the hint

Answer 38

You had to have solved a couple of DEs before to know the general form

Answer 39

Waaaaaah I don't still understand XD

Answer 40

may be lets create a table

Answer 41

Your first question complex root, along with its conjugate not repeated* sorry

Answer 42

that's alright :)

but please help me to solve my second question :(

Answer 43

ganeshie8 my second question is y=Ae^2x+Bxe^2x

Answer 44

y = Ae^2x + Bxe^2x = e^2x(A + Bx)
y' = e^2x(0+B) + 2e^2x(A+Bx) = Be^2x + 2y
y'' = 2Be^2x + 2y'
-----------------------------

Answer 45

I have just differentiated the equation twice ^

Answer 46

oh i see :) this way of elimination works too

Answer 47

we still need to eliminate B from the last equation... think of a clever way to combine those 3 equations... hmmm im still thinking..

Answer 48

Thanks for your help guys :)

Answer 49

it gets a little long but i can see a way, just start rearranging, b interms of y'' and ' and then the first equation in terms of b and x

Answer 50

yeah we can subtract last two equations i think

Answer 51

then you will have to find a a linear combination where b and A are 0 or something

Answer 52

0 : y = Ae^2x + Bxe^2x = e^2x(A + Bx)
-2 : y' = e^2x(0+B) + 2e^2x(A+Bx) = Be^2x + 2y
1 : y'' = 2Be^2x + 2y'
-----------------------------
- 2y'+y'' = - 4y + 2y'

Answer 53

nice =]

Answer 54

the answer is -4y+2y' am I right?

Answer 55

the answer is an equation, don't forget the left hand side !

Answer 56

- 2y'+y'' = - 4y + 2y'

y'' - 4y' + 4y = 0

Answer 57

ahh ok so the answer should be -2y'+y"=-4y'+2y'

Answer 58

Ok thanks ganeshie8!!!!!!!!! :)

Answer 59

yes, and you may make it look neat by putting it in standard form ^

Answer 60

ganeshie8 and dan815 Thanks for your help guys! Ajah :)

Answer 61

1 big smile for you guys! :)

Answer 62

yep also note thee are other solutions like any other multiple of the left side

Answer 63

k(y'' - 4y' + 4y) = 0
k E R

Answer 64

since they would be dependant

Answer 65

maybe I shouldnt say other solutons, but different way to say the same solution, because when you are eliminatnig, you might not always end up with the simplest representation

Answer 66

so, your answer on a test might not look the same way, so just rearrange till it does

Answer 67

dan815 ok so you rearrange again the solution?