`y^(3)dx+x^(2)dy=0`

Options

A. `x^(-1)+2y^(-2)=c`
B. `2x^(-1)+y^(-1)=c`
C. `x^(-1)+y^(-2)=2c`
D. `x^(-1)-2y^(-1)=c`

Answer

To solve the differential equation \( y^{3}dx + x^{2}dy = 0 \), we can follow these steps:

### Step 1: Rearranging the Equation
We start by rearranging the given equation:
\[ x^{2}dy = -y^{3}dx \]
Now, we can separate the variables:
\[ \frac{dy}{y^{3}} = -\frac{dx}{x^{2}} \]

### Step 2: Integrating Both Sides
Next, we integrate both sides:
\[ \int \frac{dy}{y^{3}} = \int -\frac{dx}{x^{2}} \]

The left side integrates to:
\[ \int y^{-3} dy = \frac{y^{-2}}{-2} = -\frac{1}{2y^{2}} + C_1 \]

The right side integrates to:
\[ \int -x^{-2} dx = \frac{x^{-1}}{1} = -\frac{1}{x} + C_2 \]

### Step 3: Equating the Integrals
Now we can equate the two integrals:
\[ -\frac{1}{2y^{2}} = -\frac{1}{x} + C \]
where \( C = C_2 - C_1 \) is a constant.

### Step 4: Simplifying the Equation
Multiplying through by -1 gives:
\[ \frac{1}{2y^{2}} = \frac{1}{x} - C \]
Rearranging gives:
\[ \frac{1}{2y^{2}} + \frac{1}{x} = C \]

### Step 5: Final Form
To express this in a more standard form, we can multiply through by \( 2xy^{2} \):
\[ x + 2y^{2}Cxy^{2} = 2y^{2} \]
Thus, we can write the solution as:
\[ \frac{1}{2y^{2}} + \frac{1}{x} = A \]
where \( A \) is a new constant.

### Final Solution
The solution to the differential equation is:
\[ \frac{1}{2y^{2}} + \frac{1}{x} = A \]

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