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Denebel

  • 2 years ago

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

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  1. myininaya
    • 2 years ago
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    \[e^{-y} dy =\cos(x)e^{\sin(x)} dx\]

  2. myininaya
    • 2 years ago
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    integrate both sides

  3. Denebel
    • 2 years ago
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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?

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

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

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

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

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

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

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

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

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

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

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

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

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

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

  18. Denebel
    • 2 years ago
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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?

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

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

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

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

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

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

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

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

  27. myininaya
    • 2 years ago
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    yes

  28. Denebel
    • 2 years ago
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    Thanks!

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