The equation of the curve through the point ( 1,2) and whose slope is `(y-1)/(x^(2) + x)` , is

Options

A. `(y-1) ( x+1) -2x=0`
B. `2x(y-1)+ x+ 1=0`
C. `x(y-1)(x+1)+2 =0`
D. None of these

Answer

To find the equation of the curve through the point (1, 2) with the given slope \(\frac{y - 1}{x^2 + x}\), we can follow these steps:

### Step 1: Set up the differential equation
The slope of the curve is given as:
\[ \frac{dy}{dx} = \frac{y - 1}{x^2 + x} \]

### Step 2: Separate the variables
We can rearrange the equation to separate the variables \(y\) and \(x\):
\[ \frac{dy}{y - 1} = \frac{dx}{x^2 + x} \]

### Step 3: Integrate both sides
Now we will integrate both sides. The left side integrates to:
\[ \int \frac{dy}{y - 1} = \ln |y - 1| + C_1 \]
For the right side, we can factor the denominator:
\[ \int \frac{dx}{x^2 + x} = \int \frac{dx}{x(x + 1)} \]
We can use partial fraction decomposition:
\[ \frac{1}{x(x + 1)} = \frac{A}{x} + \frac{B}{x + 1} \]
Multiplying through by the denominator:
\[ 1 = A(x + 1) + Bx \]
Setting \(x = 0\):
\[ 1 = A(0 + 1) \Rightarrow A = 1 \]
Setting \(x = -1\):
\[ 1 = B(-1) \Rightarrow B = -1 \]
Thus, we have:
\[ \int \frac{dx}{x^2 + x} = \int \left(\frac{1}{x} - \frac{1}{x + 1}\right) dx = \ln |x| - \ln |x + 1| + C_2 \]

### Step 4: Combine the results
Now we combine the results of the integrals:
\[ \ln |y - 1| = \ln |x| - \ln |x + 1| + C \]
Using the property of logarithms:
\[ \ln |y - 1| = \ln \left(\frac{x}{x + 1}\right) + C \]

### Step 5: Exponentiate both sides
Exponentiating both sides gives:
\[ |y - 1| = e^C \cdot \frac{x}{x + 1} \]
Let \(k = e^C\), we can rewrite this as:
\[ y - 1 = k \cdot \frac{x}{x + 1} \]

### Step 6: Solve for \(y\)
Thus, we have:
\[ y = k \cdot \frac{x}{x + 1} + 1 \]

### Step 7: Use the initial condition
We know the curve passes through the point (1, 2):
\[ 2 = k \cdot \frac{1}{1 + 1} + 1 \]
\[ 2 = \frac{k}{2} + 1 \]
Subtracting 1 from both sides:
\[ 1 = \frac{k}{2} \]
Multiplying both sides by 2:
\[ k = 2 \]

### Step 8: Write the final equation
Substituting \(k\) back into the equation:
\[ y = 2 \cdot \frac{x}{x + 1} + 1 \]
This simplifies to:
\[ y = \frac{2x}{x + 1} + 1 = \frac{2x + (x + 1)}{x + 1} = \frac{3x + 1}{x + 1} \]

### Final Equation
Thus, the equation of the curve is:
\[ (x + 1)(y - 1) = 2x \]