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anonymous

  • 5 years ago

Find y. y'-e^ysinx=0

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  1. bahrom7893
    • 5 years ago
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    another diff eqn?

  2. anonymous
    • 5 years ago
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    yess

  3. bahrom7893
    • 5 years ago
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    yay lol workin on it.

  4. bahrom7893
    • 5 years ago
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    FIRST OF ALL, In ALL Differential equations, especially if they are in terms of x and y, not only one variable, rewrite y' as dy/dx

  5. bahrom7893
    • 5 years ago
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    dy/dx - (e^y)(Sinx) = 0 dy/dx = (e^y)(Sinx)

  6. bahrom7893
    • 5 years ago
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    Divide both sides by e^y and multiply by dx: dy/e^y = Sinx dx

  7. bahrom7893
    • 5 years ago
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    For first integral: rewrite dy/e^y as e^(-y)dy, then let u = -y; du = -dy

  8. bahrom7893
    • 5 years ago
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    So: Integral of (e^(-y)dy) = - Integral of (e^(-y)(-dy)) [I multiplied by two minuses ( double negative is a positive) to get a -dy=du]

  9. bahrom7893
    • 5 years ago
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    - Integral of (e^(-y)(-dy)) = - Integral of (e^u du) = - e^u = - e^(-y)

  10. bahrom7893
    • 5 years ago
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    Int (dy/e^y)= Int (Sinx dx) - e^(-y) = - Cos(x) + C

  11. bahrom7893
    • 5 years ago
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    multiply everything by -1 ( - 1 times a constant is still a constant so I will let -C be A) e^(-y) = Cos(x) + A

  12. bahrom7893
    • 5 years ago
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    Take Ln of both sides: Ln(e^(-y)) = Ln(Cos(x)+A) -y = Ln(Cos(x)+A) y = - Ln(Cos(x)+A)

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