Question

how to solve y''+y-2+0.....

Answer 1

?you can't solve it because it isnt an equation

Answer 2

sorry.....its y'' + y =2

Answer 3

first lets solve y''+y=2 I think you can find e's exponent by doing r^2+1=0

Answer 4

its been awhile since it is imaginary you might use cos and sin is that right?

Answer 5

i mean first lets solve y''+y=0

Answer 6

lol oops

Answer 7

the solution of that will be part of the solution to y''+y=2

Answer 8

yc= c1 exp[-x] + c2 exp[-x] is this right

Answer 9

you can check it y'=-c1exp(-x)-c2exp(-x) and y''=c1exp(-x)+c2exp(-x) so no u have to use the sin and cos solution

Answer 10

how to right in terms of sin nd cos?

Answer 11

i think its y=c1sin[sqrt(2)t]-c2sin[sqrt(2)]

Answer 12

i missed my t in that second part

Answer 13

roots to characteristic equation are +i,-i

Answer 14

you are right i don't know where I got r^2+2

Answer 15

should had been r^2+1

Answer 16

so we replace those sqrt of 2 with just 1*t instead of sqrt(2)*t

Answer 17

we have for homogeneous solution part is y=c1cost-c2sint

Answer 18

and we can check it

Answer 19

so y'=-c1sint-c2cost and y''=-c1cost+c2cost

Answer 20

gosh i hate typing lol

Answer 21

but i have two conditions y(x=0)=0 & y(x=1)=0

Answer 22

y''=-c1cost+c2sint

Answer 23

we arent done yet we will get to the conditions

Answer 24

-c1cost+c2sint+c1cost-c2sint=0 so we are good for y''+y=0

Answer 25

now we need to do y''+y=2

Answer 26

ok....

Answer 27

how to get fr dat 2???

Answer 28

http://www.sosmath.com/tables/diffeq/diffeq.html
this is an awesome site

Answer 29

are solution for the homogenous was not the general solution it was only a particular

Answer 30

do you see where it says 2nd order

Answer 31

yep..

Answer 32

y=c1.cos(x) + c2.sin(x) is the general solution i no....bt hw 2 get particular solution

Answer 33

ok we did find the general solution since alpha=0 and Belta=1,-1 since that is the coefficent i

Answer 34

yeah i just found that out

Answer 35

so once we find the nonhomogeneous solution then we done with the solution part until we get to using initial condition's. We have to make a guess at y

Answer 36

y=0 for both x=1 & x=0 are the given initial conditions...these conditions yield 0=0!!!

Answer 37

we arent done finding the solution we cant use the intial condition yet

Answer 38

so how to find the particular solution....i no dat v hav found d general solution

Answer 39

we have only found the general solution for the homogeneous equation now we need to find the particular solution for the non homogeneous part. we have to make a guess at y

Answer 40

what if we let y=A

Answer 41

A is a constant.

Answer 42

is y=A a particular solution

Answer 43

y'= what?

Answer 44

and what does y''=?

Answer 45

wat vil b d pert soltn for dis prblm??

Answer 46

so in order for both sides to be the same A has to equal what?

Answer 47

what is y' and y'' if y=A where A is a constant

Answer 48

0

Answer 49

thank you so what does A have to be?

Answer 50

2??

Answer 51

so what is your whole solution without using the initial conditions

Answer 52

y= 2 + c1.cos x - c2.sin x

Answer 53

yes now do you know how to find the solution with initial conditions?

Answer 54

ohh...yeah thank you very much!!!!

Answer 55

ok have fun. thanks for the memory refresher

Answer 56

and one more question i have about finite element analysis....i am new so can u tell me ver 2 ask d question??