Question

I'm going over some theorems regarding the existence of a unique solution for IVP's. Not really sure what it's asking me to do.

Answer 1

"Find a member of the family that is a solution of the initial value problem."
\[y=c_{1}e^{x}+c_{2}e^{-x}, \ \ \ (-\infty,\infty)\]\[y''+y=0, y(0)=0, y'(0)=1\]

Answer 2

I believe that is just asking you to solve the IVP

Answer 3

find, c1 and c2. At least that is all that it appears to be

Answer 4

What about all the member family stuff? Yeah, I'm gonna check the answers in the back to get an idea of exactly what they are asking for, but the problem is moreso that I don't understand the theory of exactly what they are asking. My reading on this always leaves me with a pretty hazy understanding, no matter how much I look at it.

Answer 5

it looks like a system of equations

Answer 6

The family is the general solution without solved c1 and c2

Answer 7

memeber is the specific solution to this IVP

Answer 8

"The given family of functions is the general solution of the differential equation on the indicated interval."

Answer 9

yea, and?

Answer 10

So it's asking for a specific solution to an IVP when it's speaking about a member of a family of solutions?

Answer 11

(Was just posting that quote for clarity, not correcting you or anybody)

Answer 12

yep, specific sln for the specific family member

Answer 13

ah ok

Answer 14

In the second term of every y value, the exponent is supposed to be negative, right?

Answer 15

yes, because you have e^-x

Answer 16

but you know how to find the first and second deriv of y correct?

Answer 17

do you know how to solve the IVP as well?

Answer 18

Yeah, I know how to find the derivatives of y.

A little confused about perl's result, because it looks like you come up with c_{1}=c_{2}, and c_{1}=-c_{2} for the two eqns, a contradiction.

Answer 19

yea, just try it and show me your work.

Answer 20

I can tell you if it is correct or not once I see your work

Answer 21

From which point? The very beginning, the derivative, or what?

Answer 22

(I'll start from the beginning)

Answer 23

from the derivatives,

Answer 24

@perl, you typed up the question wrong when you did it.

Answer 25

oh sorry

Answer 26

\[y = c_{1}e^{x}+c_{2}e^{-x}; \ \ \ \frac{dy}{dx}=c_{1}e^{x}-c_{2}e^{-x}.\]\[\frac{d^{2}y}{dx^{2}}=c_{1}e^{x}+c^{2}e^{-x}.\]

Answer 27

although it did end up not mattering because you did the derivatives correctly

Answer 28

ok, so now, plug and chug

Answer 29

Plugging in to the original Differential Equation:

\[y''+y=0;\]\[\bigg[c_{1}e^{x}+c_{2}e^{-x}\bigg]+\bigg[c_{1}e^{x}+c_{2}e^{-x}\bigg]=0\]

Answer 30

ok, so now what does that simplify to?

Answer 31

Not sure where the squared came from, up above was typo.

Answer 32

same here, which is why I deleted it as soon as I posted it lol

Answer 33

Just two times both terms;\[2c_{1}e^{x}+2c_{2}e^{-x}=0\]

Answer 34

is there a type-o in the question?

Answer 35

Let me check.

Answer 36

y = c1e^x + c2e^(-x)

y(0) = 0 :
`0 = c1 + c2`

y'(0) = 1 :
`1 = c1 - c2`

Answer 37

Yup, sorry, folks, it's minus. y''-y=0, which does indeed equal zero, not sure about the significance of that result, anyways.

Answer 38

y = c1*e^x + c2*e^x
y' = c1*e^x - c2*e^x
y ' ' = c1 * e^x +c2^e^x


1 = y(0) = c1*e^0 + c2*e^0

0 = y'(0) = c1*e^0 - c2*e^0

1 = c1 + c2
0 = c1 - c2

1 = 2*c1

c1 = 1/2

c2 = -1/2

Answer 39

ok so now as ganeshie and perl pointed out, you need to solve the system created by the first derivative and the original y

Answer 40

Alright, so the general procedure for a problem like this with the given information-if I follow-is to take the necessary derivatives (y^(n)), and evaluate those derivatives at the given initial or boundary conditions to then solve for the constants present in the solution

Answer 41

yup, solve for your constants then check using the problem and plugging and chugging

Answer 42

same thing when you get to laplace IVPs

Answer 43

that typo in the question was significant, it threw me off :)

Answer 44

same here, I realized it after I said it

Answer 45

because we have zeroes

Answer 46

you realized there was a typo because it wasnt satisfying y ' ' + y = 0

Answer 47

the given solution y(x) was not satisfying y ' ' + y = 0, etc

Answer 48

the question was a contradiction

Answer 49

I made the fault of only looking at the first and thinking you then had the derivatives wrong as well. But yea, that's why I realized the type-o. Only solution was trivial doesn't make sense

Answer 50

right :)

Answer 51

Alright, I'm getting lost in talking about this problem as to where I move from here; in the situation that we have where the derivatives have been taken, the appropriate equations have been evaluated given the initial or boundary conditions, and the constant coefficients have been solved for, is that it? Or is there anything else to do.

Answer 52

yep that's it. You have done it

Answer 53

you can use the original to check your answer

Answer 54

You guys did it, heh. I'll have to work an example on my own, but yeah, thank you, guys.

Answer 55

you can check with wolfram (just as a final check)

http://www.wolframalpha.com/input/?i=solve+y%27+%27+-+y+%3D+0%2C+y%280%29%3D0%2C+y+%27+%280%29+%3D+1+

Answer 56

you actually did it. If you can find the derivatives

Answer 57

then once that I trust you can solve the system

Answer 58

(It's really tempting to just sort of, uh, put my head down here in the library, and fall asleep, lol...)

Answer 59

interestingly, the original question I got y = sin x as a solution
http://www.wolframalpha.com/input/?i=solve+y%27+%27+%2B+y+%3D+0%2C+y%280%29%3D0%2C+y+%27+%280%29+%3D+1+

Answer 60

lol yea? I get that it's late