The solution of the differential equation `(dy)/(dx)=-((x^(2)+3y^(2))/(3x^(2)+y^(2))),y(1)=0` is

Options

A. `log_(e)|x+y|-(xy)/((x+y)^(2))=0`
B. `log_(e)|x+y|-(2xy)/((x+y)^(2))=0`
C. `log_(e)|x+y|+(xy)/((x+y)^(2))=0`
D. `log_(e)|x+y|+(2x)/((x+y)^(2))=0`

Answer

To solve the differential equation

\[ \frac{dy}{dx} = -\frac{x^2 + 3y^2}{3x^2 + y^2}, \quad y(1) = 0, \]

we will follow these steps:

### Step 1: Identify the type of differential equation
The given equation is a first-order homogeneous differential equation because both the numerator and denominator are of degree 2.

### Step 2: Substitute \( y = vx \)
Let \( y = vx \), where \( v \) is a function of \( x \). Then, we have:

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

### Step 3: Substitute into the differential equation
Substituting \( y = vx \) into the differential equation gives:

\[ v + x \frac{dv}{dx} = -\frac{x^2 + 3(vx)^2}{3x^2 + (vx)^2}. \]

This simplifies to:

\[ v + x \frac{dv}{dx} = -\frac{x^2 + 3v^2x^2}{3x^2 + v^2x^2} = -\frac{1 + 3v^2}{3 + v^2}. \]

### Step 4: Rearranging the equation
Rearranging gives:

\[ x \frac{dv}{dx} = -\frac{1 + 3v^2}{3 + v^2} - v. \]

### Step 5: Combine terms
Combining the terms on the right-hand side:

\[ x \frac{dv}{dx} = -\frac{1 + 3v^2 + v(3 + v^2)}{3 + v^2} = -\frac{1 + 3v^2 + 3v + v^3}{3 + v^2}. \]

### Step 6: Separate variables
Now, we can separate the variables:

\[ \frac{3 + v^2}{1 + 3v^2 + 3v + v^3} dv = -\frac{1}{x} dx. \]

### Step 7: Integrate both sides
Integrating both sides gives:

\[ \int \frac{3 + v^2}{1 + 3v^2 + 3v + v^3} dv = -\int \frac{1}{x} dx. \]

### Step 8: Solve the integrals
The left side can be solved using partial fractions or substitution, while the right side integrates to:

\[ -\ln |x| + C. \]

### Step 9: Apply the initial condition
Using the initial condition \( y(1) = 0 \) implies \( v(1) = 0 \). Substitute \( x = 1 \) and \( v = 0 \) into the integrated equation to find \( C \).

### Step 10: Solve for \( y \)
After integrating and applying the initial condition, we will express \( y \) in terms of \( x \) to get the final solution.

### Final Solution
The final solution will be in the form:

\[ \ln |x + y| + \frac{2xy}{(x + y)^2} = C, \]

where \( C \) is determined from the initial condition.