The solution of the differential equation `x(dy)/(dx)=y ln ((y^(2))/(x^(2)))` is (where, c is an arbitrary constant)

Options

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

Answer

To solve the differential equation \( x \frac{dy}{dx} = y \ln \left( \frac{y^2}{x^2} \right) \), we can follow these steps:

### Step 1: Rewrite the Differential Equation
We start with the given equation:
\[ x \frac{dy}{dx} = y \ln \left( \frac{y^2}{x^2} \right) \]
This can be rewritten as:
\[ \frac{dy}{dx} = \frac{y}{x} \ln \left( \frac{y^2}{x^2} \right) \]

### Step 2: Simplify the Logarithm
Using properties of logarithms, we can simplify:
\[ \ln \left( \frac{y^2}{x^2} \right) = \ln(y^2) - \ln(x^2) = 2 \ln(y) - 2 \ln(x) \]
Thus, the equation becomes:
\[ \frac{dy}{dx} = \frac{y}{x} (2 \ln(y) - 2 \ln(x)) \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{y}{x} (2 \ln(y) - 2 \ln(x)) = \frac{2y}{x} (\ln(y) - \ln(x)) \]

### Step 3: Homogeneous Function Substitution
We can use the substitution \( y = vx \), where \( v \) is a function of \( x \). Then, we have:
\[ \frac{dy}{dx} = v + x \frac{dv}{dx} \]
Substituting \( y = vx \) into the equation gives:
\[ v + x \frac{dv}{dx} = \frac{vx}{x} (2 \ln(vx) - 2 \ln(x)) \]
This simplifies to:
\[ v + x \frac{dv}{dx} = v (2 \ln(v) + 2 \ln(x) - 2 \ln(x)) = 2v \ln(v) \]
Thus, we have:
\[ x \frac{dv}{dx} = 2v \ln(v) - v \]
Factoring out \( v \):
\[ x \frac{dv}{dx} = v(2 \ln(v) - 1) \]

### Step 4: Separate Variables
We can separate the variables:
\[ \frac{dv}{v(2 \ln(v) - 1)} = \frac{dx}{x} \]

### Step 5: Integrate Both Sides
Integrating both sides:
\[ \int \frac{dv}{v(2 \ln(v) - 1)} = \int \frac{dx}{x} \]
The right side integrates to \( \ln|x| + C_1 \).

### Step 6: Solve the Left Side Integral
For the left side, we can use substitution \( t = \ln(v) \), which gives \( dv = v dt = e^t dt \):
\[ \int \frac{e^t dt}{e^t(2t - 1)} = \int \frac{dt}{2t - 1} \]
This integrates to:
\[ \frac{1}{2} \ln|2t - 1| + C_2 \]
Substituting back \( t = \ln(v) \):
\[ \frac{1}{2} \ln|2 \ln(v) - 1| + C_2 = \ln|x| + C_1 \]

### Step 7: Exponentiate to Solve for \( v \)
Exponentiating both sides gives:
\[ |2 \ln(v) - 1| = C |x|^2 \]
where \( C = e^{2C_2} \).

### Step 8: Substitute Back for \( y \)
Substituting back \( v = \frac{y}{x} \):
\[ |2 \ln\left(\frac{y}{x}\right) - 1| = C |x|^2 \]
This can be rearranged to find \( y \):
\[ \ln\left(\frac{y}{x}\right) = \frac{1 + Cx^2}{2} \]
Thus, we find:
\[ y = x e^{\frac{1 + Cx^2}{2}} \]

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