(c) x(e

ସ௬ − 1) ௗ௬

ௗ௫

+ (x

ଶ − 1)e

ଶ௬ = 0, x > 0 solve using separation of variables 

We are given the differential equation:

\[
x e^{y} \frac{dy}{dx} + (x^2 - 1)e^{2y} = 0, \quad x > 0
\]

We will solve this using the method of separation of variables.

### Step 1: Rewriting the equation
We start by rewriting the equation:

\[
x e^{y} \frac{dy}{dx} = -(x^2 - 1) e^{2y}
\]

Now, divide both sides by \( e^y \):

\[
x \frac{dy}{dx} = -(x^2 - 1) e^y
\]

### Step 2: Separate the variables
We want to separate the variables \( x \) and \( y \). So, divide both sides by \( -(x^2 - 1) \) and multiply by \( dx \):

\[
\frac{dy}{e^y} = - \frac{(x^2 - 1)}{x} dx
\]

### Step 3: Integrate both sides
Now we can integrate both sides. The left side requires integrating with respect to \( y \), and the right side with respect to \( x \).

For the left side:

\[
\int \frac{dy}{e^y} = \int - \frac{(x^2 - 1)}{x} dx
\]

The left-hand side simplifies to:

\[
\int e^{-y} dy = -e^{-y}
\]

For the right-hand side, we simplify the integrand:

\[
\int - \frac{(x^2 - 1)}{x} dx = \int -x + \frac{1}{x} dx
\]

Now integrate:

\[
- \int x dx + \int \frac{1}{x} dx = - \frac{x^2}{2} + \ln |x|
\]

### Step 4: Combine the results
We now combine both results from the integrals:

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

Where \( C \) is the constant of integration. Simplifying, we get:

\[
e^{-y} = \frac{x^2}{2} - \ln |x| - C
\]

### Step 5: Solve for \( y \)
Now, solve for \( y \) by taking the natural logarithm of both sides:

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

Thus, the solution to the differential equation is:

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

### Summary
Using separation of variables, we have solved the differential equation and obtained the general solution for \( y \) in terms of \( x \).

Would you like more details on any of the steps?

Here are some related questions to deepen your understanding:

1. How can separation of variables be applied to other types of differential equations?
2. What is the importance of the constant of integration in solving differential equations?
3. How do you handle boundary conditions or initial conditions when solving differential equations?
4. Why is it necessary to check the domain of the solution after solving?
5. Can the method of separation of variables be extended to partial differential equations?

**Tip**: Always check the consistency of units or dimensions when solving physical problems involving differential equations.

(c) x(e^{y} − 1) \frac{dy}{dx} + (x^2 − 1)e^{2y} = 0, x > 0 solve using separation of variables

Discover how to solve the differential equation x e^y dy/dx + (x^2 - 1) e^{2y} = 0 using the separation of variables method. The solution involves key calculus concepts such as integration techniques and manipulation of exponential functions. Suitable for college-level students.

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