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X(x-1) y’ + y(y-1) using separating differencialne rovnice

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Answer to a math question X(x-1) y’ + y(y-1) using separating differencialne rovnice

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Fred
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To solve the differential equation X(x-1)y' + y(y-1) = 0 using separation of variables, we first rewrite it in the form:

X(x-1)y' + y(y-1) = 0

\frac{X(x-1)}{y(1-y)} dy = -dx

Now, we will separate variables and integrate:

\int \frac{X(x-1)}{y(1-y)} dy = \int -dx

\int \frac{1}{y(1-y)} dy = -\int \frac{1}{X(x-1)} dx

\int \left(\frac{1}{y} + \frac{1}{1-y}\right) dy = - \frac{1}{X} \int \frac{1}{x-1} dx

\int \left(\frac{1}{y} + \frac{1}{1-y}\right) dy = - \frac{1}{X} \ln|x-1| + C

Integrate with respect to y:

\ln|y| - \ln|1-y| = - \frac{1}{X} \ln|x-1| + C

Simplify and rewrite in terms of y:

\ln\left|\frac{y}{1-y}\right| = - \frac{1}{X} \ln|x-1| + C

Exponentiate both sides to eliminate the logarithms:

\left|\frac{y}{1-y}\right| = C (x-1)^{-X}

Solving for y:

\frac{y}{1-y} = \pm C (x-1)^{-X}

y = \frac{C(x-1)^{-X}}{1 \pm C(x-1)^{-X}}

Therefore, the solution to the differential equation is:

\boxed{y = \frac{C(x-1)^{-X}}{1 \pm C(x-1)^{-X}}}

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