`(y-1)sinx dx - dy = 0` Solve the first-order differential equation

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`(y-1)sin(x)dx - dy = 0`

To solve, express the equation in the form `N(y)dy = M(x)dx`.

So bringing same variables on one side, the equation becomes:

`(y-1) sin(x) dx = dy`

`sin(x) dx = dy/(y - 1)`

Then, take the integral of both sides.

`int sin(x) dx = int dy/(y-1)`

For the left side, apply the formula `int sin (u) du = -cos(u) + C` .

And for the right side, apply the formula `int (du)/u =ln|u| + C` .

`-cos(x) +C_1 = ln|y-1|+C_2`

From here, isolate the y.

`-cos(x) + C_1 - C_2 = ln|y-1|`

Since C1 and C2 represent any number, express it as a single constant C.

`-cos(x) +C = ln|y-1|`

Then, eliminate the logarithm in the equation.

`e^(-cos(x)+C) = e^(ln|y-1|)`

`e^(-cos(x) + C) = |y-1|`

`+-e^(-cos(x) + C) = y-1`

To simplify the left side, apply the exponent rule `a^m*a^n=a^(m+n)` .

`+-e^(-cos(x))*e^C= y-1`


Since `+-e^C` is a constant, it can be replaced with C.

`Ce^(-cos(x))=y - 1`


Therefore, the general solution is  `y=Ce^(-cos(x))+1` .

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