Solution of `y dx – x dy = x^2 ydx` is:

Options

A. `ye ^(x^2)= cx^2`
B. `ye ^(-x^2)= cx^2`
C. `y^2e^(x^2) = cx^2`
D. `y^2 e^(-x^2)= cx^2`

Answer

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

### Step 1: Rearranging the Equation
We start with the given equation:
\[ y \, dx - x \, dy = x^2 y \, dx \]
We can rearrange this to isolate terms involving \( dx \) and \( dy \):
\[ y \, dx - x^2 y \, dx = x \, dy \]
Factoring out \( dx \) from the left side gives:
\[ y(1 - x^2) \, dx = x \, dy \]

### Step 2: Separating Variables
Next, we can separate the variables \( x \) and \( y \):
\[ \frac{dy}{y} = \frac{1 - x^2}{x} \, dx \]

### Step 3: Integrating Both Sides
Now, we integrate both sides:
\[ \int \frac{dy}{y} = \int \frac{1 - x^2}{x} \, dx \]
The left side integrates to:
\[ \ln |y| + C_1 \]
For the right side, we can split the integral:
\[ \int \left( \frac{1}{x} - x \right) \, dx = \ln |x| - \frac{x^2}{2} + C_2 \]
Thus, we have:
\[ \ln |y| = \ln |x| - \frac{x^2}{2} + C \]
where \( C = C_2 - C_1 \).

### Step 4: Exponentiating Both Sides
Exponentiating both sides to eliminate the logarithm gives:
\[ |y| = e^{\ln |x| - \frac{x^2}{2} + C} = |x| e^{C} e^{-\frac{x^2}{2}} \]
Let \( K = e^{C} \), then:
\[ y = K x e^{-\frac{x^2}{2}} \]

### Step 5: Rearranging the Equation
We can rearrange this equation to match the form of the options given:
\[ y e^{\frac{x^2}{2}} = K x \]
Squaring both sides gives:
\[ y^2 e^{x^2} = K^2 x^2 \]
Letting \( K^2 = c \), we have:
\[ y^2 e^{x^2} = c x^2 \]

### Final Result
Thus, the solution to the differential equation is:
\[ y^2 e^{x^2} = c x^2 \]

### Conclusion
The correct option is:
**Third option: \( y^2 e^{x^2} = c x^2 \)**.