The solution of the differential equation
` x^(2) (dy)/(dx)= x^(2) + xy + y^(2) `

Options

A. ` tan^(-1)(y/x)=2 log x+C`
B. ` tan^(-1)(y/x)=3 log x+C`
C. ` tan^(-1)(y/x)= log x +C`
D. ` tan^(-1)(y/x)= 4 log x+C`

Explanation

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

### Step 1: Rearranging the equation
We start by rewriting the given differential equation:

\[ \frac{dy}{dx} = \frac{x^2 + xy + y^2}{x^2} \]

### Step 2: Substituting \( y = vx \)
Next, we will use the substitution \( y = vx \), where \( v \) is a function of \( x \). This gives us:

\[ \frac{dy}{dx} = v + x \frac{dv}{dx} \]

### Step 3: Substitute into the equation
Substituting \( y = vx \) into the rearranged equation, we have:

\[ v + x \frac{dv}{dx} = \frac{x^2 + x(vx) + (vx)^2}{x^2} \]

This simplifies to:

\[ v + x \frac{dv}{dx} = \frac{x^2 + vx^2 + v^2x^2}{x^2} \]

### Step 4: Simplifying the right-hand side
Now, we simplify the right-hand side:

\[ v + x \frac{dv}{dx} = 1 + v + v^2 \]

### Step 5: Rearranging the equation
Subtract \( v \) from both sides:

\[ x \frac{dv}{dx} = 1 + v^2 \]

### Step 6: Separating variables
Now, we separate the variables:

\[ \frac{dv}{1 + v^2} = \frac{dx}{x} \]

### Step 7: Integrating both sides
We integrate both sides:

\[ \int \frac{dv}{1 + v^2} = \int \frac{dx}{x} \]

The left side integrates to \( \tan^{-1}(v) \) and the right side integrates to \( \ln|x| + C \):

\[ \tan^{-1}(v) = \ln|x| + C \]

### Step 8: Back-substituting for \( v \)
Recall that \( v = \frac{y}{x} \). Thus, we have:

\[ \tan^{-1}\left(\frac{y}{x}\right) = \ln|x| + C \]

### Final Solution
The final solution to the differential equation is:

\[ \tan^{-1}\left(\frac{y}{x}\right) = \ln|x| + C \]

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