Question

(ODE) Dealing with a straightforward IVP, am given initial conditions *and* a family of functions that is the general solution to the IVP, just need to find a member of the family that is a solution as well.

Answer 1

http://i.imgur.com/rTlwRkY.png

Answer 2

(Upper bound is infinity, some characters don't display right in that PDF)

Answer 3

should be straightforward yeah

Answer 4

Alright, so I have the family of solutions, now I think I just need to like, take derivatives and plug in or something and solve for constants

Answer 5

Is that right?

Answer 6

ye

Answer 7

Alright, lol, now the difference between saying it and doing it, going to take a shot now, but in general I'm pretty bumbly/not-good with remembering this stuff in practice. One sec.

Answer 8

Taking the derivatives:

Answer 9

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

Answer 10

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

Answer 11

y'' is not needed

Answer 12

Plugging in the constants/conditions:

Answer 13

Oh, okay, uh, why not?

Answer 14

use the initial conditions for y and y'

Answer 15

OH

Answer 16

you will get two equations
solve the constants

Answer 17

Yeah, I misread, I'm trying to go quick and misread it as an IVP condition

Answer 18

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

Answer 19

(I'm going about this in the wrong way, I see what I need to do)

Answer 20

Just add the equations together without rearranging.

Answer 21

\[y(0) = 0 \implies c_1+c_2 = 0\tag{1}\]
\[y'(0) = 1 \implies c_1-c_2 = 1\tag{2}\]

Answer 22

Oh, because x is zero, yeah, but works the same way, add them together, eliminate c_{2}, 2c_{1}=1, c_{1}= 1/2

Answer 23

\[2c_{1}=1; \ \ \ c_{1}=\frac{1}{2}\]

Answer 24

\[\frac{1}{2}-c_{2}=1; \ \ \ c_{2}=-\frac{1}{2}\]

Answer 25

Lemme check book answers

Answer 26

Yup, that's right, now moving on to a different type of problem, one minute.