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

- anonymous

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

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- myininaya

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

- anonymous

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

- myininaya

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

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## More answers

- myininaya

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

- myininaya

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

- myininaya

lol oops

- myininaya

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

- anonymous

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

- myininaya

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

- anonymous

how to right in terms of sin nd cos?

- myininaya

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

- myininaya

i missed my t in that second part

- anonymous

roots to characteristic equation are +i,-i

- myininaya

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

- myininaya

should had been r^2+1

- myininaya

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

- myininaya

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

- myininaya

and we can check it

- myininaya

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

- myininaya

gosh i hate typing lol

- anonymous

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

- myininaya

y''=-c1cost+c2sint

- myininaya

we arent done yet we will get to the conditions

- myininaya

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

- myininaya

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

- anonymous

ok....

- anonymous

how to get fr dat 2???

- myininaya

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

- myininaya

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

- myininaya

do you see where it says 2nd order

- anonymous

yep..

- anonymous

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

- myininaya

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

- myininaya

yeah i just found that out

- myininaya

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

- anonymous

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

- myininaya

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

- anonymous

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

- myininaya

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

- myininaya

what if we let y=A

- myininaya

A is a constant.

- myininaya

is y=A a particular solution

- myininaya

y'= what?

- myininaya

and what does y''=?

- anonymous

wat vil b d pert soltn for dis prblm??

- myininaya

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

- myininaya

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

- anonymous

0

- myininaya

thank you so what does A have to be?

- anonymous

2??

- myininaya

so what is your whole solution without using the initial conditions

- anonymous

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

- myininaya

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

- anonymous

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

- myininaya

ok have fun. thanks for the memory refresher

- anonymous

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

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