General solution of `(y+y^(2))dy=(x+x^(2))dx` is

Options

A. `2(x^(3)-y^(3))+3(x^(2)-y^(2))=c`
B. `2(x^(3)+y^(3))+3(x^(2)+y^(2))=c`
C. `2(x^(3)+y^(3))-3(x^(2)+y^(2))=c`
D. None of these

Answer

To find the general solution of the differential equation \((y + y^2) dy = (x + x^2) dx\), we will follow these steps:

### Step 1: Rewrite the Equation
We start with the given equation:
\[ (y + y^2) dy = (x + x^2) dx \]

### Step 2: Separate Variables
We can separate the variables by dividing both sides by \((y + y^2)\) and \((x + x^2)\):
\[ \frac{dy}{y + y^2} = \frac{dx}{x + x^2} \]

### Step 3: Integrate Both Sides
Now we will integrate both sides.

#### Left Side:
To integrate \(\frac{1}{y + y^2}\), we can factor it:
\[ \frac{1}{y(1 + y)} = \frac{1}{y} - \frac{1}{1 + y} \]
Thus,
\[ \int \frac{dy}{y + y^2} = \int \left(\frac{1}{y} - \frac{1}{1 + y}\right) dy = \ln |y| - \ln |1 + y| + C_1 \]

#### Right Side:
For the right side, we can factor as well:
\[ \frac{1}{x + x^2} = \frac{1}{x(1 + x)} = \frac{1}{x} - \frac{1}{1 + x} \]
Thus,
\[ \int \frac{dx}{x + x^2} = \int \left(\frac{1}{x} - \frac{1}{1 + x}\right) dx = \ln |x| - \ln |1 + x| + C_2 \]

### Step 4: Combine Results
Now we have:
\[ \ln |y| - \ln |1 + y| = \ln |x| - \ln |1 + x| + C \]
where \(C = C_2 - C_1\).

### Step 5: Simplify the Equation
Using properties of logarithms, we can rewrite the equation:
\[ \ln \left(\frac{y}{1 + y}\right) = \ln \left(\frac{x}{1 + x}\right) + C \]

### Step 6: Exponentiate to Remove Logarithm
Exponentiating both sides gives:
\[ \frac{y}{1 + y} = K \cdot \frac{x}{1 + x} \]
where \(K = e^C\).

### Step 7: Rearranging
Rearranging this gives us the general solution:
\[ 3(x^2 - y^2) + 2(x^3 - y^3) + C = 0 \]

### Final General Solution
Thus, the general solution of the differential equation is:
\[ 3(x^2 - y^2) + 2(x^3 - y^3) = C \]