The solution of the equation `(dy)/(dx) =(y^(2)-y-2)/(x^(2)+2x-3)` is

Options

A. `(1)/(3)log|(y-2)/(y+1)|=(1)/(4)log|(x-1)/(x+3)|+c`
B. `(1)/(3)log|(y+1)/(y-2)|=(1)/(4)log|(x+3)/(x-1)|+c`
C. `(1)/(4)log|(y+1)/(y-2)| =(1)/(3) log|(x+3)/(x-1)|+c`
D. None of these

Answer

To solve the differential equation

\[ \frac{dy}{dx} = \frac{y^2 - y - 2}{x^2 + 2x - 3}, \]

we will use the method of separation of variables.

### Step 1: Factor the numerator and denominator

First, we need to factor both the numerator and the denominator.

**Numerator:**
\[ y^2 - y - 2 = (y - 2)(y + 1). \]

**Denominator:**
\[ x^2 + 2x - 3 = (x + 3)(x - 1). \]

So, we can rewrite the equation as:

\[ \frac{dy}{dx} = \frac{(y - 2)(y + 1)}{(x + 3)(x - 1)}. \]

### Step 2: Separate the variables

Now, we can separate the variables:

\[ \frac{dy}{(y - 2)(y + 1)} = \frac{dx}{(x + 3)(x - 1)}. \]

### Step 3: Integrate both sides

Next, we integrate both sides. We will perform partial fraction decomposition on both sides.

**Left Side:**

For \(\frac{1}{(y - 2)(y + 1)}\), we can write:

\[ \frac{1}{(y - 2)(y + 1)} = \frac{A}{y - 2} + \frac{B}{y + 1}. \]

Multiplying through by the denominator \((y - 2)(y + 1)\) gives:

\[ 1 = A(y + 1) + B(y - 2). \]

Setting \(y = 2\):

\[ 1 = A(2 + 1) \implies A = \frac{1}{3}. \]

Setting \(y = -1\):

\[ 1 = B(-1 - 2) \implies B = -\frac{1}{3}. \]

Thus, we have:

\[ \frac{1}{(y - 2)(y + 1)} = \frac{1/3}{y - 2} - \frac{1/3}{y + 1}. \]

Integrating gives:

\[ \int \left( \frac{1/3}{y - 2} - \frac{1/3}{y + 1} \right) dy = \frac{1}{3} \ln |y - 2| - \frac{1}{3} \ln |y + 1| + C_1. \]

**Right Side:**

For \(\frac{1}{(x + 3)(x - 1)}\), we can write:

\[ \frac{1}{(x + 3)(x - 1)} = \frac{C}{x + 3} + \frac{D}{x - 1}. \]

Multiplying through by the denominator gives:

\[ 1 = C(x - 1) + D(x + 3). \]

Setting \(x = -3\):

\[ 1 = D(-3 + 3) \implies D = \frac{1}{4}. \]

Setting \(x = 1\):

\[ 1 = C(1 - 1) + D(1 + 3) \implies C = \frac{1}{4}. \]

Thus, we have:

\[ \frac{1}{(x + 3)(x - 1)} = \frac{1/4}{x + 3} - \frac{1/4}{x - 1}. \]

Integrating gives:

\[ \int \left( \frac{1/4}{x + 3} - \frac{1/4}{x - 1} \right) dx = \frac{1}{4} \ln |x + 3| - \frac{1}{4} \ln |x - 1| + C_2. \]

### Step 4: Combine the results

Setting the integrals equal to each other gives:

\[ \frac{1}{3} \ln |y - 2| - \frac{1}{3} \ln |y + 1| = \frac{1}{4} \ln |x + 3| - \frac{1}{4} \ln |x - 1| + C. \]

### Step 5: Simplify

This can be simplified to:

\[ \ln \left( \frac{|y - 2|^{1/3}}{|y + 1|^{1/3}} \right) = \ln \left( \frac{|x + 3|^{1/4}}{|x - 1|^{1/4}} \right) + C. \]

Exponentiating both sides gives:

\[ \frac{|y - 2|^{1/3}}{|y + 1|^{1/3}} = K \cdot \frac{|x + 3|^{1/4}}{|x - 1|^{1/4}}, \]

where \(K = e^C\).

### Final Result

Thus, the general solution of the differential equation is:

\[ \frac{|y - 2|^{1/3}}{|y + 1|^{1/3}} = K \cdot \frac{|x + 3|^{1/4}}{|x - 1|^{1/4}}. \]