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ksaimouli

  • 2 years ago

solve

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  1. ksaimouli
    • 2 years ago
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    \[\frac{ dy }{ dx }=x^2e^{4x}\]

  2. ksaimouli
    • 2 years ago
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    solve the differential equation

  3. ksaimouli
    • 2 years ago
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    i tried using by parts

  4. satellite73
    • 2 years ago
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    i think that just means find the anti derivative of \(x^2e^{4x}\)and my guess is use parts twice

  5. ksaimouli
    • 2 years ago
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    ya i used

  6. ksaimouli
    • 2 years ago
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    but not solved let me show

  7. ksaimouli
    • 2 years ago
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    i will use 7 technique

  8. ksaimouli
    • 2 years ago
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    |dw:1360895129811:dw|

  9. ksaimouli
    • 2 years ago
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    |dw:1360895181802:dw|

  10. ksaimouli
    • 2 years ago
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    |dw:1360895191642:dw|

  11. satellite73
    • 2 years ago
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    i think you have it backwards you want to drop the power on the \(x^2\) term

  12. ksaimouli
    • 2 years ago
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    and now i should choose x^3 and e^4x

  13. satellite73
    • 2 years ago
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    no the gimmick is to reduce the power, not raise it. raising the power makes it worse

  14. satellite73
    • 2 years ago
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    \(u=x^2, dv = e^{4x}dx, du = 2x, v =\frac{1}{4}e^{4x}\)

  15. ksaimouli
    • 2 years ago
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    okay i can do it this way so something wrong in my technique right

  16. satellite73
    • 2 years ago
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    yes. as you can see your method just gave higher and higher powers of \(x\) you want to make them lower and lower

  17. ksaimouli
    • 2 years ago
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    so instead of integrating id i take derivative

  18. ksaimouli
    • 2 years ago
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    but if i follow the LIPET rule it says u= log right

  19. satellite73
    • 2 years ago
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    i forget exactly how that box method works, but it is \[\int u dv =uv-\int v du\]

  20. satellite73
    • 2 years ago
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    make \(u=x^2, du = 2xdx, dv = e^{4x}dx, v =\frac{1}{4}e^{4x}\) get \[\int x^2e^{4x}=\frac{1}{4}e^{4x}x^2-\frac{1}{2}\int e^{4x}xdx\] then do it again

  21. ksaimouli
    • 2 years ago
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    okay got u

  22. ksaimouli
    • 2 years ago
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    thx

  23. satellite73
    • 2 years ago
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    yw

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