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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).

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  1. anonymous
    • 5 years ago
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    Obviously, y is going to be e as is y'.

  2. anonymous
    • 5 years ago
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    the DE equation solution is y(x)=e^x

  3. anonymous
    • 5 years ago
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    The value of y(e) = e^e.

  4. anonymous
    • 5 years ago
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    I believe the answer is e^e, but I dont know the steps to get there

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

  6. anonymous
    • 5 years ago
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    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 .-)

  7. anonymous
    • 5 years ago
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    yes, but they don't give steps, I get confused whenever y'=y type questions come up

  8. myininaya
    • 5 years ago
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    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

  9. myininaya
    • 5 years ago
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    y(x)=0 is found after applying the initial condition

  10. anonymous
    • 5 years ago
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    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?

  11. myininaya
    • 5 years ago
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    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 =

  12. myininaya
    • 5 years ago
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    oops y''-5y'+6y=0

  13. anonymous
    • 5 years ago
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    oh ok, so c1e^rx is what we use with differentials when we solve them like quadratics?

  14. myininaya
    • 5 years ago
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    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)

  15. myininaya
    • 5 years ago
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    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

  16. myininaya
    • 5 years ago
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    along with the e thing

  17. anonymous
    • 5 years ago
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    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

  18. anonymous
    • 5 years ago
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    ok thanks a lot, you've really helped out

  19. myininaya
    • 5 years ago
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    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)

  20. myininaya
    • 5 years ago
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    thanks andra

  21. anonymous
    • 5 years ago
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    no prob, I will have my exam about this topic in a week

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