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

(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
+-e^C*e^(-cos(x))=y-1
Since +-e^C is a constant, it can be replaced with C.
Ce^(-cos(x))=y - 1
Ce^(-cos(x))+1=y
Therefore, the general solution is  y=Ce^(-cos(x))+1 .