What is the solution for (dy)/(dx)=(1-2x)/(4x-x^2) ?

The solution for (dy)/(dx)=(1-2x)/(4x-x^2) is y= 1/4 ln(x/(x-4))-2(-ln(x-4)-1)+c_{1}

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(dy)/(dx)=(1-2x)/(4x-x^2)

\frac{d y}{d x} = \frac{1 - 2 x}{4 x - x ^{2}}

y = \frac{1}{4} ln (\frac{x}{x - 4}) - 2 (- ln (x - 4) - 1) + c_{1}

Solution steps

\frac{d y}{d x} = \frac{1 - 2 x}{4 x - x ^{2}}

Substitute

\frac{d y}{d x}

with

y^{'} (x)

y^{^{'}} (x) = \frac{1 - 2 x}{4 x - x ^{2}}

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

y = \int \frac{1 - 2 x}{4 x - x ^{2}} d x

\int \frac{1 - 2 x}{4 x - x ^{2}} d x = \frac{1}{4} (ln (\frac{x - 2}{2} + 1) - ln (\frac{x - 2}{2} - 1)) - 2 (\frac{x}{4 - x} - ln (x - 4) + \frac{4}{x - 4}) + c_{1}

y = \frac{1}{4} (ln (\frac{x - 2}{2} + 1) - ln (\frac{x - 2}{2} - 1)) - 2 (\frac{x}{4 - x} - ln (x - 4) + \frac{4}{x - 4}) + c_{1}

Simplify

y = \frac{1}{4} ln (\frac{x}{x - 4}) - 2 (- ln (x - 4) - 1) + c_{1}

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What is the solution for (dy)/(dx)=(1-2x)/(4x-x^2) ?
The solution for (dy)/(dx)=(1-2x)/(4x-x^2) is y= 1/4 ln(x/(x-4))-2(-ln(x-4)-1)+c_{1}