Answer:
To solve the first-order differential equation \((x - 1)y' = y - 2\), you can use the method of separation of variables. Here's how you can do it:
1. Start by rearranging the equation to isolate the variables \(x\) and \(y\) on opposite sides of the equation:
\((x - 1)y' = y - 2\)
2. Divide both sides by \((x - 1)\) to get the \(y'\) term on one side:
\[y' = \frac{y - 2}{x - 1}\]
3. Now, we will separate the variables by moving all the \(y\) terms to one side and all the \(x\) terms to the other side. This involves multiplying both sides by \((x - 1)\) and dividing both sides by \((y - 2)\):
\[\frac{1}{y - 2} dy = \frac{1}{x - 1} dx\]
4. Integrate both sides with respect to their respective variables:
\[\int \frac{1}{y - 2} dy = \int \frac{1}{x - 1} dx\]
5. The left-hand side can be integrated using a simple substitution. Let \(u = y - 2\), then \(du = dy\):
\[\int \frac{1}{u} du = \int \frac{1}{x - 1} dx\]
Now, integrate both sides:
\[\ln|u| = \ln|x - 1| + C_1\]
Where \(C_1\) is the constant of integration.
6. Remove the absolute value on the left-hand side by taking the absolute value of both sides:
\[\ln|y - 2| = \ln|x - 1| + C_1\]
7. Now, you can simplify the equation by removing the natural logarithms:
\[|y - 2| = |x - 1|e^{C_1}\]
8. Since \(e^{C_1}\) is just another constant, let's denote it as \(C_2\):
\[|y - 2| = C_2|x - 1|\]
9. Finally, remove the absolute value bars by considering two cases:
Case 1: \(y - 2 = C_2(x - 1)\)
Case 2: \(y - 2 = -C_2(x - 1)\)
These two cases cover all possibilities.
So, the solution to the differential equation \((x - 1)y' = y - 2\) is given by:
\[y = C_2(x - 1) + 2\]
or
\[y = -C_2(x - 1) + 2\]
Where \(C_2\) is an arbitrary constant that depends on the initial conditions of the problem.
Explanation: