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Denebel Group Title

use separation of variables to solve the initial value problem. dy/dx=(cos x)e^(y+sinx) and y=0 when x=0

  • 2 years ago
  • 2 years ago

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  1. myininaya Group Title
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    \[e^{-y} dy =\cos(x)e^{\sin(x)} dx\]

    • 2 years ago
  2. myininaya Group Title
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    integrate both sides

    • 2 years ago
  3. Denebel Group Title
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    Lol.. I don't know how to move "the x with dx" and "y with dy".. :/ can you show me that part?

    • 2 years ago
  4. myininaya Group Title
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    i multiplied dx on both sides

    • 2 years ago
  5. myininaya Group Title
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    \[e^{y+\sin(x)}=e^{y} e^{\sin(x)}\]

    • 2 years ago
  6. myininaya Group Title
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    divide both sides by e^y

    • 2 years ago
  7. myininaya Group Title
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    1/e^{y}=e^{-y}

    • 2 years ago
  8. Denebel Group Title
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    Oh ok. Thanks.

    • 2 years ago
  9. Denebel Group Title
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    How do I integrate cosxe^(sin x) dx?

    • 2 years ago
  10. myininaya Group Title
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    ley u=sin(x) => du=cos(x) dx

    • 2 years ago
  11. Denebel Group Title
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    ? What about the e? Do I put e to the u power?

    • 2 years ago
  12. myininaya Group Title
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    \[\int\limits_{}^{}e^u du=e^u+C=e^{\sin(x)}+C\]

    • 2 years ago
  13. myininaya Group Title
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    i replace sin(x) with u and cos(x) dx with du

    • 2 years ago
  14. Denebel Group Title
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    Oh? But don't I integrate |dw:1327549907891:dw|? What happens with the cos...

    • 2 years ago
  15. myininaya Group Title
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    cos(x) dx=du i replaced cos(x) dx with du i did integrate

    • 2 years ago
  16. myininaya Group Title
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    |dw:1327550585485:dw|

    • 2 years ago
  17. myininaya Group Title
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    |dw:1327550603285:dw|

    • 2 years ago
  18. Denebel Group Title
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    Ohh I see. Ok, what do I do now? Do I plug in the given x, y values to find C?

    • 2 years ago
  19. myininaya Group Title
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    have you integrated the other side yet?

    • 2 years ago
  20. Denebel Group Title
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    I have now: -e^(-y) = e^(sinx) + C

    • 2 years ago
  21. myininaya Group Title
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    ok so enter in (0,0) and solve for C

    • 2 years ago
  22. Denebel Group Title
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    I solved that C = 0, So does the equation become -e^(-y) = e^(sinx) ?

    • 2 years ago
  23. myininaya Group Title
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    -1-1=-2

    • 2 years ago
  24. myininaya Group Title
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    -1=1+C -1-1=C -2=C

    • 2 years ago
  25. Denebel Group Title
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    Oh I forgot about the negative on the left.

    • 2 years ago
  26. Denebel Group Title
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    Then -e^(-y) = e^(sinx) - 2 ?

    • 2 years ago
  27. myininaya Group Title
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    yes

    • 2 years ago
  28. Denebel Group Title
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    Thanks!

    • 2 years ago
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