If x=x(y) is the solution of the differential equation, `ydx-(x+2y^2)dy=0`, with `x(-pi)=pi^2` , then x is equal to :

Options

A. `-2y^2+3piy`
B. `2y^2+piy`
C. `2y^2-pi^2`
D. `-y^2-2piy`

Answer

To solve the differential equation \( y \, dx - (x + 2y^2) \, dy = 0 \) with the initial condition \( x(-\pi) = \pi^2 \), we will follow these steps:

### Step 1: Rearranging the Differential Equation
We start with the given differential equation:
\[ y \, dx - (x + 2y^2) \, dy = 0 \]
Rearranging gives:
\[ y \, dx = (x + 2y^2) \, dy \]
Dividing both sides by \( y \):
\[ dx = \left( \frac{x}{y} + 2y \right) dy \]

### Step 2: Separating Variables
We can rewrite the equation as:
\[ \frac{dx}{dy} = \frac{x}{y} + 2y \]
This is a first-order linear differential equation in the form \( \frac{dx}{dy} - \frac{x}{y} = 2y \).

### Step 3: Finding the Integrating Factor
The integrating factor \( \mu(y) \) is given by:
\[ \mu(y) = e^{\int -\frac{1}{y} \, dy} = e^{-\ln|y|} = \frac{1}{|y|} \]
We can ignore the absolute value since \( y \) is negative in our case.

### Step 4: Multiplying by the Integrating Factor
Multiplying the entire equation by \( \frac{1}{y} \):
\[ \frac{1}{y} \frac{dx}{dy} - \frac{x}{y^2} = 2 \]

### Step 5: Integrating Both Sides
The left-hand side can be rewritten as:
\[ \frac{d}{dy}\left(\frac{x}{y}\right) = 2 \]
Integrating both sides gives:
\[ \frac{x}{y} = 2y + C \]
Thus,
\[ x = 2y^2 + Cy \]

### Step 6: Applying the Initial Condition
We have the initial condition \( x(-\pi) = \pi^2 \):
\[ \pi^2 = 2(-\pi)^2 + C(-\pi) \]
Calculating this gives:
\[ \pi^2 = 2\pi^2 - C\pi \]
Rearranging gives:
\[ C\pi = 2\pi^2 - \pi^2 = \pi^2 \implies C = \pi \]

### Step 7: Final Expression for \( x \)
Substituting \( C \) back into the equation for \( x \):
\[ x = 2y^2 + \pi y \]

### Conclusion
Thus, the solution for \( x \) is:
\[ \boxed{2y^2 + \pi y} \]