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?you can't solve it because it isnt an equation
sorry.....its y'' + y =2
first lets solve y''+y=2 I think you can find e's exponent by doing r^2+1=0
its been awhile since it is imaginary you might use cos and sin is that right?
i mean first lets solve y''+y=0
the solution of that will be part of the solution to y''+y=2
yc= c1 exp[-x] + c2 exp[-x] is this right
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
how to right in terms of sin nd cos?
i think its y=c1sin[sqrt(2)t]-c2sin[sqrt(2)]
i missed my t in that second part
roots to characteristic equation are +i,-i
you are right i don't know where I got r^2+2
should had been r^2+1
so we replace those sqrt of 2 with just 1*t instead of sqrt(2)*t
we have for homogeneous solution part is y=c1cost-c2sint
and we can check it
so y'=-c1sint-c2cost and y''=-c1cost+c2cost
gosh i hate typing lol
but i have two conditions y(x=0)=0 & y(x=1)=0
we arent done yet we will get to the conditions
-c1cost+c2sint+c1cost-c2sint=0 so we are good for y''+y=0
now we need to do y''+y=2
how to get fr dat 2???
http://www.sosmath.com/tables/diffeq/diffeq.html this is an awesome site
are solution for the homogenous was not the general solution it was only a particular
do you see where it says 2nd order
y=c1.cos(x) + c2.sin(x) is the general solution i no....bt hw 2 get particular solution
ok we did find the general solution since alpha=0 and Belta=1,-1 since that is the coefficent i
yeah i just found that out
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
y=0 for both x=1 & x=0 are the given initial conditions...these conditions yield 0=0!!!
we arent done finding the solution we cant use the intial condition yet
so how to find the particular solution....i no dat v hav found d general solution
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
what if we let y=A
A is a constant.
is y=A a particular solution
and what does y''=?
wat vil b d pert soltn for dis prblm??
so in order for both sides to be the same A has to equal what?
what is y' and y'' if y=A where A is a constant
thank you so what does A have to be?
so what is your whole solution without using the initial conditions
y= 2 + c1.cos x - c2.sin x
yes now do you know how to find the solution with initial conditions?
ohh...yeah thank you very much!!!!
ok have fun. thanks for the memory refresher
and one more question i have about finite element analysis....i am new so can u tell me ver 2 ask d question??