Question

Hey everyone, solve the differential equation y'=y subject to the initial condition y(0)=0. find the value of y(e).

Answer 1

Obviously, y is going to be e as is y'.

Answer 2

the DE equation solution is y(x)=e^x

Answer 3

The value of y(e) = e^e.

Answer 4

I believe the answer is e^e, but I dont know the steps to get there

Answer 5

I am unsure. It has been a while since Diff Eq. Sorry I can't help more. Have you tried Wolfram Alpha?

Answer 6

the DE equation solution is y(x)=e^x
that is only one solution!
the general one is Ae^x
initial: y(0)=0 implies 0=Ae^0=A
so the solution for this is y=0.... boring .-)

Answer 7

yes, but they don't give steps, I get confused whenever y'=y type questions come up

Answer 8

y'=y
y'-y=0
r-1=0 implies r=1
so the solution is y=ce^x
y(0)=ce^0=c=0
so y=0
so y(e)=0

Answer 9

y(x)=0 is found after applying the initial condition

Answer 10

ok thanks, I don't understand where the e comes from I'll have to look over the notes better, but the answer is y(e)= 0?

Answer 11

hey mathty
if we have r^2-5r+6=0, the solutions are r=2 and r=3
so if we have y''-5y'+6=0
then the solution is y(x)=c1*e^(2x)+c2*e^(3x)
you can check this but plugging in and see if both sides =

Answer 12

oops y''-5y'+6y=0

Answer 13

oh ok, so c1e^rx is what we use with differentials when we solve them like quadratics?

Answer 14

if we have r^2-2r+1=0, the solutions are r=1
so if we have y''-2y'+y=0
then the solution is y(x)=ce^(x)

Answer 15

right but when you start solving these and you get imaginary answers to your quadratic
your solution will have sinx and cosx in it
here is a very helpful link
http://www.sosmath.com/diffeq/second/second.html

Answer 16

along with the e thing

Answer 17

if we have r^2-2r+1=0, the solutions are r=1 so if we have y''-2y'+y=0 then the solution is y(x)=ce^(x)
this is not correct
if there is one solution than y(x)=(a+bx)e^x

Answer 18

ok thanks a lot, you've really helped out

Answer 19

http://www.sosmath.com/diffeq/second/constantcof/constantcof.html
andras was right i did leave something out of that solution and this website points it out
the answer will be y(x)=c1e^(x)+c2xe*(x)

Answer 20

thanks andra

Answer 21

no prob, I will have my exam about this topic in a week