Differential Equation help ~

solve the differential equation sin x y'-2y cos x =0

I want to know the steps, the answer is c sin^2 x, c is a constant

sin x y'-2y cos x =0

so y' * (sinx) = 2y cos x

so y' = 2y cos x /sinx = 2y cotx

so dy/dx = 2y cotx

so dy/y = 2 dx cotx

integrate both sides

log(y) = 2 log|sin x| + k

so log(y) = log(sin^2 x) + k

so log(y) = log(sin^2 x) + log(c)

so log(y) = log{c(sin^2 x)}

taking antilog y = c sin^2 x

k can be written as log(c) for some value of c
i have done this to make things simple
or else u can see this way

so log(y) = log(sin^2 x) + k
so take anti log
u will get
y = e^k * sin^2 x
and now let e^k = c

hope its clear now
cheers

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y'=dy/dx so DE is sinxdy/dx-2ycosx=0. Arrange with x, dx together and y, dy
together. dy/y=2(cosx/sinx) dx
Now integrate: lny=2ln(sinx)+k where k is a constant
2lnsinx=ln(sin^2(x)) so lny=ln(sin^2(x)+lnc where c is a constant since k is a constant.
Giving lny=ln(csin^2(x)) and so y=csin^2(x) as required.

Isn't it separable?

sin(x)y' - 2cos(x)y = 0
sin(x)y' = 2cos(x)y
y'/y = 2cos(x)/sin(x)
dy/y = 2cos(x)/sin(x) dx
ln(y) = 2 du/u
ln(y) = 2 ln(sin(x)) + c
y = e^(2 ln(sin(x) + c)
y = c*e^(2 ln(sin(x))
y = c*sin²(x)

... sin(x) ( dy/dx) - 2y cos(x) =0
or ∫ (1/y) dy = 2 ∫ cot(x) dx
or ln(y) = 2 ln(sin(x)) + constant
or ln(y) = ln(sin²(x)) + ln(c)
or ln(y) = ln(c sin²(x))
or y = c sin²(x)