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anonymous
 5 years ago
how to solve y''+y2+0.....
anonymous
 5 years ago
how to solve y''+y2+0.....

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myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0?you can't solve it because it isnt an equation

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry.....its y'' + y =2

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0first lets solve y''+y=2 I think you can find e's exponent by doing r^2+1=0

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0its been awhile since it is imaginary you might use cos and sin is that right?

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0i mean first lets solve y''+y=0

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0the solution of that will be part of the solution to y''+y=2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yc= c1 exp[x] + c2 exp[x] is this right

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0you 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
 5 years ago
Best ResponseYou've already chosen the best response.0how to right in terms of sin nd cos?

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0i think its y=c1sin[sqrt(2)t]c2sin[sqrt(2)]

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0i missed my t in that second part

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0roots to characteristic equation are +i,i

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0you are right i don't know where I got r^2+2

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0should had been r^2+1

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0so we replace those sqrt of 2 with just 1*t instead of sqrt(2)*t

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0we have for homogeneous solution part is y=c1costc2sint

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0so y'=c1sintc2cost and y''=c1cost+c2cost

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0gosh i hate typing lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but i have two conditions y(x=0)=0 & y(x=1)=0

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0we arent done yet we will get to the conditions

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0c1cost+c2sint+c1costc2sint=0 so we are good for y''+y=0

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0now we need to do y''+y=2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how to get fr dat 2???

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0http://www.sosmath.com/tables/diffeq/diffeq.html this is an awesome site

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0are solution for the homogenous was not the general solution it was only a particular

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0do you see where it says 2nd order

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0y=c1.cos(x) + c2.sin(x) is the general solution i no....bt hw 2 get particular solution

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0ok we did find the general solution since alpha=0 and Belta=1,1 since that is the coefficent i

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0yeah i just found that out

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0so 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
 5 years ago
Best ResponseYou've already chosen the best response.0y=0 for both x=1 & x=0 are the given initial conditions...these conditions yield 0=0!!!

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0we arent done finding the solution we cant use the intial condition yet

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so how to find the particular solution....i no dat v hav found d general solution

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0we 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
 5 years ago
Best ResponseYou've already chosen the best response.0is y=A a particular solution

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wat vil b d pert soltn for dis prblm??

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0so in order for both sides to be the same A has to equal what?

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0what is y' and y'' if y=A where A is a constant

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0thank you so what does A have to be?

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0so what is your whole solution without using the initial conditions

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0y= 2 + c1.cos x  c2.sin x

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0yes now do you know how to find the solution with initial conditions?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ohh...yeah thank you very much!!!!

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0ok have fun. thanks for the memory refresher

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and 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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