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-3)|= (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. `log (y^(2)-y-2) = log (x^(2) + 2x-3)+c`

Answer

To solve the differential equation

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

we will follow these steps:

### Step 1: Factor the numerator and denominator
First, we need to factor both the numerator \(y^2 - y - 2\) and the denominator \(x^2 + 2x - 3\).

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

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

Thus, we can rewrite the equation as:
\[ \frac{dy}{dx} = \frac{(y - 2)(y + 1)}{(x + 3)(x - 1)} \]

### Step 2: Separate the variables
We can separate the variables by rearranging the equation:
\[ \frac{dy}{(y - 2)(y + 1)} = \frac{dx}{(x + 3)(x - 1)} \]

### Step 3: Integrate both sides
Next, we will integrate both sides.

**Left Side:**
To integrate \(\frac{1}{(y - 2)(y + 1)}\), we can use partial fractions:
\[ \frac{1}{(y - 2)(y + 1)} = \frac{A}{y - 2} + \frac{B}{y + 1} \]

Multiplying through by the denominator \((y - 2)(y + 1)\) gives:
\[ 1 = A(y + 1) + B(y - 2) \]

Setting \(y = 2\):
\[ 1 = A(3) \implies A = \frac{1}{3} \]

Setting \(y = -1\):
\[ 1 = B(-3) \implies B = -\frac{1}{3} \]

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

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 the right side, we can similarly use partial fractions:
\[ \frac{1}{(x + 3)(x - 1)} = \frac{C}{x + 3} + \frac{D}{x - 1} \]

Following similar steps, we find:
\[ \int \left( \frac{1/4}{x + 3} - \frac{1/4}{x - 1} \right) dx = \frac{1}{4} \ln |x + 3| - \frac{1}{4} \ln |x - 1| + C_2 \]

### Step 4: Combine results
Setting the integrals equal to each other:
\[ \frac{1}{3} \ln |y - 2| - \frac{1}{3} \ln |y + 1| = \frac{1}{4} \ln |x + 3| - \frac{1}{4} \ln |x - 1| + C \]

### Step 5: Simplify
We can combine the logarithms:
\[ \frac{1}{3} \ln \left( \frac{|y - 2|}{|y + 1|} \right) = \frac{1}{4} \ln \left( \frac{|x + 3|}{|x - 1|} \right) + C \]

### Step 6: Exponentiate to solve for y
Exponentiating both sides gives:
\[ \left( \frac{|y - 2|}{|y + 1|} \right)^{\frac{1}{3}} = K \left( \frac{|x + 3|}{|x - 1|} \right)^{\frac{1}{4}} \]
where \(K = e^{3C}\).

### Final Solution
Thus, the solution of the differential equation is:
\[ \frac{|y - 2|}{|y + 1|} = K \left( \frac{|x + 3|}{|x - 1|} \right)^{\frac{3}{4}}. \]