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Mathematics 13 Online
OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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 .-)

OpenStudy (anonymous):

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

myininaya (myininaya):

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

myininaya (myininaya):

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

OpenStudy (anonymous):

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?

myininaya (myininaya):

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 =

myininaya (myininaya):

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

OpenStudy (anonymous):

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

myininaya (myininaya):

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)

myininaya (myininaya):

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

myininaya (myininaya):

along with the e thing

OpenStudy (anonymous):

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

OpenStudy (anonymous):

ok thanks a lot, you've really helped out

myininaya (myininaya):

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)

myininaya (myininaya):

thanks andra

OpenStudy (anonymous):

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

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