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anonymous

  • 5 years ago

does anyone understand Differential equations

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  1. lalaly
    • 5 years ago
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    yes

  2. amistre64
    • 5 years ago
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    einstein did i think

  3. amistre64
    • 5 years ago
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    newton faked it lol

  4. Mr.Math
    • 5 years ago
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    lol

  5. anonymous
    • 5 years ago
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    i just begun D.E. and i am totally lost. find a function, y=f(x) satisfying the given differential equation and the prescribed initial condition. dy/dx= 2x+1; y(0)=3 where do i start

  6. amistre64
    • 5 years ago
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    with integration id assume

  7. lalaly
    • 5 years ago
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    multiply both sides by dx .. then integrate both sides

  8. amistre64
    • 5 years ago
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    \[y' = 2x+1\] \[\int y' = \int 2x+1\] \[y=x^2+x +C\]

  9. amistre64
    • 5 years ago
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    plug in x=0 and y=3 to solve for C

  10. lalaly
    • 5 years ago
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    this is a seperable DE

  11. amistre64
    • 5 years ago
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    yeah, semantics :)

  12. lalaly
    • 5 years ago
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    :D:D

  13. anonymous
    • 5 years ago
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    you do these just like seperable this is 2 sections before seperable

  14. lalaly
    • 5 years ago
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    Initial Value problem?

  15. anonymous
    • 5 years ago
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    intergrals as general and particular solutions

  16. lalaly
    • 5 years ago
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    forget what i said, its done the way amistre showd u

  17. amistre64
    • 5 years ago
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    yay!!

  18. lalaly
    • 5 years ago
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    hehehe

  19. anonymous
    • 5 years ago
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    ok appreciate it!

  20. amistre64
    • 5 years ago
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    \[Y(x)=\int y'\ dx\] \[Y(x)=\int (2x+1)dx\] stuff like that

  21. anonymous
    • 5 years ago
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    it just shows you the equation and intial condition and you have to choose what to do. So when you work it out it becomes what you are doing with the integral

  22. lalaly
    • 5 years ago
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    \[\frac{dy}{dx}=2x+1\]\[dy=(2x+1)dx\]now integrate both sides\[\int\limits{dy}=\int\limits{(2x+1)dx}\]\[y=x^2+x+c\]

  23. anonymous
    • 5 years ago
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    yea, just like that, thanks,

  24. lalaly
    • 5 years ago
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    :)

  25. anonymous
    • 5 years ago
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    i need help with intergals, it has been 2 yrs since i have done them. x times sq. root of x^2+9

  26. lalaly
    • 5 years ago
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    let u=x^2+9 du=2x so integration becomes\[\frac{1}{2} \int\limits{ \sqrt u}du\]

  27. lalaly
    • 5 years ago
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    integrate sqrt u then substitute the x back

  28. lalaly
    • 5 years ago
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    du=2xdx **

  29. lalaly
    • 5 years ago
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    I am really sorry, i have to leave now. Post a new question and someone else could help you

  30. anonymous
    • 5 years ago
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    thanks

  31. anonymous
    • 5 years ago
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    is \[^{?} \int\limits_{?}^{?} xe ^{-x}\] u substitution or by parts integral

  32. amistre64
    • 5 years ago
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    or simply recognize that your missing a -1 in front

  33. amistre64
    • 5 years ago
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    i missed read it

  34. amistre64
    • 5 years ago
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    by parts with that one

  35. anonymous
    • 5 years ago
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    alright

  36. amistre64
    • 5 years ago
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    e^-x + x -e^-x | -1 e^-x | = x e^-x - e^-x +0 -----

  37. amistre64
    • 5 years ago
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    +C if youaint got bounds

  38. amistre64
    • 5 years ago
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    i missed a negative on there .... bad fingers, bad!!

  39. amistre64
    • 5 years ago
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    \[-xe^{-x}-e^{-x}\ or\ -e^{-x}(x+1)\]

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