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Obviously, y is going to be e as is y'.
the DE equation solution is y(x)=e^x
The value of y(e) = e^e.
I believe the answer is e^e, but I dont know the steps to get there
I am unsure. It has been a while since Diff Eq. Sorry I can't help more. Have you tried Wolfram Alpha?
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 .-)
yes, but they don't give steps, I get confused whenever y'=y type questions come up
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
y(x)=0 is found after applying the initial condition
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?
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 =
oh ok, so c1e^rx is what we use with differentials when we solve them like quadratics?
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)
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
along with the e thing
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
ok thanks a lot, you've really helped out
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)
no prob, I will have my exam about this topic in a week