Question

Verify that for any C the function y= (Ce^t)+1+t is a solution of the differential equation y'= y-t. Determine the value of C so that the solution satisfies the initial condition y(2)=1

Answer 1

so you just need to plug in

Answer 2

first do you know how to find y' given y=Ce^t+1+t

Answer 3

C= =2/e^2

Answer 4

-2*

Answer 5

oh I guess you are passed the verification

Answer 6

you just want to find C when t=2 and y=1

Answer 7

\[y=Ce^{t}+1+t \\ 1=Ce^{2}+1+2 \\ 1=Ce^{2}+3 \\ -2 =Ce^{2}\]
yep you are right C is -2/e^2

Answer 8

so y(t)= (-2/e^2)e^t+1+t

Answer 9

yep

Answer 10

theres a second part to it, but im not sure why i had to find that last part

Answer 11

what's the second part?

Answer 12

Using the file ode.xls determine approximately how many subintervals are needed using Euler method and the modified Euler method in order to estimate to three decimal places the solution at t=4 of the initial value problem in problem a).

Answer 13

what does it mean use the file ode.xls

Answer 14

what would be the initial problem in the previous one? I have to use excel for this, but i am having trouble figuring out the initial value

Answer 15

and i dont see how y(t)= (-2/e^2)e^t+1+t has anything to do with b

Answer 16

how many steps do we use

Answer 17

oh that is what we need to determine to get withint 3 decimal places

Answer 18

I don't know how to do that
I can do Euler formula given the step number
but I don't know how to determine the amount of steps to get within a certain accuracy

Answer 19

I will do some further investigation
I have food though right now so I will be back in a few hours
http://math.sci.ccny.cuny.edu/document/show/2156
this site is what I will be reading on my food break

Answer 20

hey back

Answer 21

so using our
\[y(t)=\frac{-2}{e^2}e^{t}+1+t \\ \text{ we evaluate this at } t=4 \\ y(4)=\frac{-2}{e^2}e^{4}+1+4 =-2e^2+5 \approx -9.778\]
now you just use the file you have to determine the number of steps to get to -9.778
using f(t,y) is y-t
enter in the exact solution -2e^2+5
initial time was given as y(2)=1 which is t=2 , 2 would be what you type in
finial time since we are looking for y(4) would be t=4, 4 would be what you type in
initial y value is 1 given from y(1)=2
number of sub-intervals is the number you need to play with

Answer 22

oops the exact solution is suppose to be y evaluated at t
that is the exact solution is (-2/e^2 )e^t+1+t

Answer 23

anyways let me know if you need more help

Answer 24

im sorry, this site froze

Answer 25

i plug in (-2/e^2)e^t+1+t into excel?

Answer 26

tell me does your ode editor thing... look like figure 3.3 http://math.sci.ccny.cuny.edu/document/show/2156

Answer 27

somewhat, the problem i am looking at is 15b

Answer 28

I do not have access to that file...
but if your "ode editor" looks like that you can enter...
f(t,y) is y-t
check the box saying you know the exact solution which we found to be (-2/e^2)e^t+1+t
then initial time was given as 2
finial time is given as 4
initial y is given as 1
and just play with the number of sub-intervals til we get -9.778

Answer 29

the exact solution would be -9.778, so just fill that in for exact solution?

Answer 30

no

Answer 31

fill in what I said

Answer 32

also, when i play with the subintervals, i go past 400...

Answer 33

they are talking about the solution to the differential equation in that field

Answer 34

oh, alright. I will fill that out

Answer 35

if you want me to play with this file I would need you to send it to me

Answer 36

i hope you dont mind

Answer 37

i will save my work, and if its okay with you, i'll send it over

Answer 38

err it is complaining about macro being diabled or whatever
and I try to enable it and it doesn't work

Answer 39

you typed in stuff right
we are suppose to look at the yapprox column
and try to get its last row as -9.778
so instead of 50 see what 100 gives

Answer 40

ok so just keep going until i hit -9.778. 100 gives me -9.489

Answer 41

double 100
try 200

Answer 42

and yes

Answer 43

keep doubling until you get there

Answer 44

okay, thank you very much

Answer 45

im up to 1000

Answer 46

where has the y approx landed there

Answer 47

-9.749

Answer 48

well that is a little closer i wonder if 2000 is going to put us over

Answer 49

-9.763

Answer 50

keep going?

Answer 51

yep double 2000

Answer 52

i did this the first time but i thought i was doing it wrong because it took so long

Answer 53

at 8000 its -9.774

Answer 54

lol
thank god you don't have to it by hand

Answer 55

we are getting closer

Answer 56

16000 is probably going to put us over... but try it and see

Answer 57

im at 20,000......

Answer 58

20,000 what does it say?

Answer 59

-9.777

Answer 60

then i have to do 40,000
and i get -9.777

Answer 61

thats good enough right?

Answer 62

i got the program to work

Answer 63

yay, this is insane. 40,000 subintervals

Answer 64

i think we stop here. it wont let me go to 80,000

Answer 65

hey....

Answer 66

exactly what number were you looking at?

Answer 67

you were suppose to be looking at the last line of the second column

Answer 68

...yes i was

Answer 69

y approx

Answer 70

thats right?

Answer 71

did i do it wrong

Answer 72

for 50 I get -9.213
for 100 I get -9.489
for 200 I get -9.632
for 400 I get -9.705
for 800 I get -9.741
for 1600 I get -9.760
for 3200 I get -9.769
for 6400 I get -9.773
for 12800 I get -9.776


.... this program is kinda weird

Answer 73

I'm trying to do 25600 right now but it goes over

Answer 74

:( my professor told me once we hit 800 we go to 1000, 2000, 4000, 8000

Answer 75

he was going over it in class, and if 20,000 is -9.777 and then 40,000 is -9.777 then its fine...but i dont know

Answer 76

it looks like the biggest number we can put in is 18,000
but if you put in a bigger number than 18,000 it is automatically going to spit out the answer for 50,000

Answer 77

and if n=50,000 it does give -9.778

Answer 78

so maybe we should go for whatever n first gave you -9.777

Answer 79

i will leave it at n=20,000

Answer 80

i emailed my professor, this is due 2 weeks from now haha

Answer 81

the problem is that the results I'm getting for n=20,000 is actually for n=50,000

Answer 82

I know this because the last row is on 50,019

Answer 83

anyways interesting program
thanks for giving it to me

Answer 84

for n=15000, I get -9.778

Answer 85

I wonder if I can find a smaller n that gives me that value

Answer 86

n=14000 also gives me -9.778

Answer 87

wait so i cant use n=20000?

Answer 88

well the problem with n=20000 as I said earlier is it is giving you the results for n=50000

Answer 89

because when i type in 20,000 for subinterval, and go down to 4, i get -9.777

Answer 90

I know but it is on row 50019

Answer 91

which means it did n=50000

Answer 92

not n=20000

Answer 93

oh, nobody in class mentioned that...

Answer 94

and i had my own modified euler instead of euler for n=14,000,18,000 whatever I entered a few posts ago

Answer 95

mentioned what?

Answer 96

oh, the problem asks for modified euler as well