Consider the differential equation `ydx-(x+y^(2))dy=0`. If for `y=1, x` takes value 1, then value of x when y = 4, is

Options

A. 64
B. 9
C. 16
D. 36

Explanation

To solve the differential equation \( y \, dx - (x + y^2) \, dy = 0 \) and find the value of \( x \) when \( y = 4 \), we can follow these steps:

### Step 1: Rewrite the Differential Equation
We start with the given differential equation:
\[ y \, dx - (x + y^2) \, dy = 0 \]
Rearranging gives:
\[ y \, dx = (x + y^2) \, dy \]

### Step 2: Separate Variables
Dividing both sides by \( y(x + y^2) \) leads to:
\[ \frac{dx}{x + y^2} = \frac{dy}{y} \]

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

### Step 4: Exponentiate to Remove Logarithm
Exponentiating both sides gives:
\[ |x + y^2| = e^C |y| \]
Let \( k = e^C \), then:
\[ x + y^2 = k y \]
or
\[ x = k y - y^2 \]

### Step 5: Find the Constant \( k \)
We use the initial condition given in the problem: when \( y = 1 \), \( x = 1 \):
\[ 1 = k(1) - (1)^2 \]
This simplifies to:
\[ 1 = k - 1 \implies k = 2 \]

### Step 6: Substitute \( k \) Back into the Equation
Substituting \( k \) back gives:
\[ x = 2y - y^2 \]

### Step 7: Find \( x \) When \( y = 4 \)
Now we substitute \( y = 4 \):
\[ x = 2(4) - (4)^2 = 8 - 16 = -8 \]

### Final Answer
Thus, the value of \( x \) when \( y = 4 \) is:
\[ \boxed{-8} \]