The solution ` x^2( dy )/( dx) = x^2 + xy + y^2` is :

Options

A. `tan ^(-1)"" (y)/(x) = bg y+C`
B. ` tan^(-1) "" (x)/(y) = log x +C `
C. ` tan ^(-1)"" (x)/(y) = log + C`
D. ` tan^(-1) "" (y)/(x) = log x+C`

Answer

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

### Step 1: Rewrite the equation
We start with the given equation:
\[ x^2 \frac{dy}{dx} = x^2 + xy + y^2 \]
We can express \(\frac{dy}{dx}\) as:
\[ \frac{dy}{dx} = \frac{x^2 + xy + y^2}{x^2} \]

### Step 2: Simplify the right-hand side
Now, we can simplify the right-hand side:
\[ \frac{dy}{dx} = 1 + \frac{y}{x} + \frac{y^2}{x^2} \]

### Step 3: Substitute \(y = vx\)
Let \(y = vx\), where \(v\) is a function of \(x\). Then, we have:
\[ \frac{dy}{dx} = v + x \frac{dv}{dx} \]
Substituting this into our equation gives:
\[ v + x \frac{dv}{dx} = 1 + v + v^2 \]

### Step 4: Simplify the equation
We can cancel \(v\) from both sides:
\[ x \frac{dv}{dx} = 1 + v^2 \]

### Step 5: Separate variables
Now, we can separate the variables:
\[ \frac{dv}{1 + v^2} = \frac{dx}{x} \]

### Step 6: Integrate both sides
Integrating 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 7: Substitute back for \(v\)
Since \(v = \frac{y}{x}\), we substitute back:
\[ \tan^{-1}\left(\frac{y}{x}\right) = \ln|x| + C \]

### Final Step: Write the final solution
Thus, the solution to the differential equation is:
\[ \tan^{-1}\left(\frac{y}{x}\right) = \ln|x| + C \]