A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 5 years ago
Hey everyone, solve the differential equation y'=y subject to the initial condition y(0)=0. find the value of y(e).
anonymous
 5 years ago
Hey everyone, solve the differential equation y'=y subject to the initial condition y(0)=0. find the value of y(e).

This Question is Closed

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Obviously, y is going to be e as is y'.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the DE equation solution is y(x)=e^x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The value of y(e) = e^e.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I believe the answer is e^e, but I dont know the steps to get there

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I am unsure. It has been a while since Diff Eq. Sorry I can't help more. Have you tried Wolfram Alpha?

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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes, but they don't give steps, I get confused whenever y'=y type questions come up

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0y'=y y'y=0 r1=0 implies r=1 so the solution is y=ce^x y(0)=ce^0=c=0 so y=0 so y(e)=0

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0y(x)=0 is found after applying the initial condition

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

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0hey mathty if we have r^25r+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 =

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh ok, so c1e^rx is what we use with differentials when we solve them like quadratics?

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0if we have r^22r+1=0, the solutions are r=1 so if we have y''2y'+y=0 then the solution is y(x)=ce^(x)

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0right 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

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0along with the e thing

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if we have r^22r+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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok thanks a lot, you've really helped out

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0http://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)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no prob, I will have my exam about this topic in a week
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.