Find y. y'-e^ysinx=0

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Find y. y'-e^ysinx=0

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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another diff eqn?
yess
yay lol workin on it.

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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
dy/dx - (e^y)(Sinx) = 0 dy/dx = (e^y)(Sinx)
Divide both sides by e^y and multiply by dx: dy/e^y = Sinx dx
For first integral: rewrite dy/e^y as e^(-y)dy, then let u = -y; du = -dy
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]
- Integral of (e^(-y)(-dy)) = - Integral of (e^u du) = - e^u = - e^(-y)
Int (dy/e^y)= Int (Sinx dx) - e^(-y) = - Cos(x) + C
multiply everything by -1 ( - 1 times a constant is still a constant so I will let -C be A) e^(-y) = Cos(x) + A
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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