Calculus of a Single Variable, Chapter 5, 5.8, Section 5.8, Problem 90

dy/dx = (1-2x) / (4x-x^2)
This differential equation is separable since it can be written in the form

N(y)dy =M(x)dx
Bringing together same variables on one side, the equation becomes
dy=(1-2x)/(4x-x^2)dx
Taking the integral of both sides, it turns into
int dy = int(1-2x)/(4x-x^2)dx
y+C_1 = int(1-2x)/(4x-x^2)dx
y+C_1= int (1-2x)/(x(4-x))dx
To take the integral of right side, apply partial fraction decomposition.
(1-2x)/(x(4-x)) = A/x + B/(4-x)

1-2x = A(4-x)+Bx

Let x=0.


1-2(0) = A(4-0)+B(0)
1=4A
1/4=A
Let x=4.
1-2(4)=A(4-4)+B(4)
-7=4B
-7/4=B

1/(2x-x^2) = (1/4)/x + (-7/4)/(4-x)

1/(2x-x^2) = 1/(4x) + (-7)/(4(4-x))

1/(2x-x^2) = 1/(4x) + (-7)/(-4(x-4))

1/(2x-x^2) = 1/(4x) + (7)/(4(x-4))
So the integrand at the right side decomposes to
y + C_1 = int (1/(4x) + 7/(4(x-4)))dx
Then, apply the formula int 1/u du = ln|u| + C .
y + C_1 = 1/4ln|x| + 7/4ln|x-4|+C_2
Isolating the y, the equation becomes
y= 1/4ln|x| + 7/4ln|x-4|+C_2-C_1
Since C1 and C2 represent any number, it can be expressed as a single constant C.
y = 1/4ln|x| + 7/4ln|x-4|+C

Therefore, the general solution of the given differential equation is y = 1/4ln|x| + 7/4ln|x-4|+C .