`(x^(2)+x)dx+(y^(2)-y)dy=0`

Options

A. `2(x^(3)+y^(3))+3(x^(2)-y^(2))=c`
B. `x^(3)+x^(2)+y^(3)+y^(3)-y^(2)=c`
C. `3(x^(3)+y^(3))+2(x^(2)-y^(2))=c`
D. `(x^(3)+y^(3))+(x^(2)-y^(2))=c`

Answer

To solve the differential equation \((x^{2}+x)dx+(y^{2}-y)dy=0\), we can follow these steps:

### Step 1: Rearranging the Equation
We start by rearranging the equation to isolate the differentials:
\[ (y^{2} - y)dy = -(x^{2} + x)dx \]

### Step 2: Integrating Both Sides
Next, we integrate both sides. We will integrate the left side with respect to \(y\) and the right side with respect to \(x\):
\[ \int (y^{2} - y) dy = -\int (x^{2} + x) dx \]

### Step 3: Performing the Integrations
Now we perform the integrations:
- For the left side:
\[ \int (y^{2} - y) dy = \frac{y^{3}}{3} - \frac{y^{2}}{2} + C_1 \]
- For the right side:
\[ -\int (x^{2} + x) dx = -\left(\frac{x^{3}}{3} + \frac{x^{2}}{2}\right) + C_2 = -\frac{x^{3}}{3} - \frac{x^{2}}{2} + C_2 \]

### Step 4: Equating the Results
Now we equate the results from both integrations:
\[ \frac{y^{3}}{3} - \frac{y^{2}}{2} = -\frac{x^{3}}{3} - \frac{x^{2}}{2} + C \]
where \(C = C_2 - C_1\).

### Step 5: Rearranging the Equation
To simplify, we can rearrange the equation:
\[ \frac{y^{3}}{3} - \frac{y^{2}}{2} + \frac{x^{3}}{3} + \frac{x^{2}}{2} = C \]

### Step 6: Multiplying Through by 6
To eliminate the fractions, we can multiply the entire equation by 6:
\[ 2y^{3} - 3y^{2} + 2x^{3} + 3x^{2} = 6C \]

### Step 7: Final Form
Letting \(A = 6C\), we can express the solution as:
\[ 2y^{3} - 3y^{2} + 2x^{3} + 3x^{2} = A \]

This is the general solution of the given differential equation.

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