Can you solve this problem please:

(a) Suppose y(x) satisfies
(yx−x+y−1)dy =1 dx
for x ≥ 0.
i. Use separation of variables to find a general solution to the ODE with
one arbitrary constant. [2]

We are given the first-order differential equation:

\[
(yx - x + y - 1) \frac{dy}{dx} = 1
\]

We need to solve this using **separation of variables**.

### Step 1: Simplify the equation
First, let’s distribute and simplify the terms on the left-hand side of the equation:

\[
\frac{dy}{dx} = \frac{1}{yx - x + y - 1}
\]

Now, rearrange the terms in the denominator for convenience:

\[
yx - x + y - 1 = y(x + 1) - (x + 1) = (y - 1)(x + 1)
\]

Thus, the differential equation becomes:

\[
\frac{dy}{dx} = \frac{1}{(y - 1)(x + 1)}
\]

### Step 2: Use separation of variables
We can now separate the variables \(x\) and \(y\):

\[
(y - 1) \, dy = \frac{1}{x + 1} \, dx
\]

### Step 3: Integrate both sides
Now, integrate both sides:

\[
\int (y - 1) \, dy = \int \frac{1}{x + 1} \, dx
\]

The integrals are straightforward:

\[
\frac{(y - 1)^2}{2} = \ln|x + 1| + C
\]

where \(C\) is the constant of integration.

### Step 4: Solve for \(y(x)\)
To express \(y\) in terms of \(x\), multiply both sides by 2:

\[
(y - 1)^2 = 2\ln|x + 1| + 2C
\]

Let \(C_1 = 2C\), then:

\[
(y - 1)^2 = 2\ln|x + 1| + C_1
\]

Finally, take the square root of both sides:

\[
y - 1 = \pm \sqrt{2\ln|x + 1| + C_1}
\]

So the general solution is:

\[
y(x) = 1 \pm \sqrt{2\ln|x + 1| + C_1}
\]

where \(C_1\) is an arbitrary constant.

### Final Answer:

\[
y(x) = 1 \pm \sqrt{2\ln|x + 1| + C_1}
\]

---

Do you want further details or have any questions?

Here are 5 related questions to expand your understanding:
1. What is the significance of the arbitrary constant \(C_1\) in the general solution?
2. How can we determine a particular solution if given an initial condition?
3. What happens to the solution as \(x\) approaches infinity?
4. Can we interpret the plus-minus sign in a physical or geometric context?
5. How do we handle cases where \(x < -1\) in this equation?

**Tip**: In separation of variables, always ensure the variables \(x\) and \(y\) are completely isolated on different sides before integrating.

Can you solve this problem please: (a) Suppose y(x) satisfies (yx−x+y−1)dy = 1 dx for x ≥ 0. i. Use separation of variables to find a general solution to the ODE with one arbitrary constant.

Discover how to solve the first-order ordinary differential equation (ODE) (yx − x + y − 1)dy = 1 dx using the separation of variables method. This solution includes step-by-step integration and covers key mathematical concepts, including the general solution y(x) = 1 ± √(2ln|x + 1| + C1). Ideal for undergraduate students.

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