Show that each of the following differential equations is homogeneous and solve each of them :
`2xy dy-(x^(2)+3y^(2))dx=0`.

Answer

To show that the differential equation \( 2xy \, dy - (x^2 + 3y^2) \, dx = 0 \) is homogeneous and to solve it, we can follow these steps:

### Step 1: Rewrite the Differential Equation
We start by rewriting the given differential equation in the standard form:
\[ 2xy \, dy = (x^2 + 3y^2) \, dx \]
Dividing both sides by \( dx \), we have:
\[ \frac{dy}{dx} = \frac{x^2 + 3y^2}{2xy} \]

### Step 2: Check for Homogeneity
A differential equation is homogeneous if the right-hand side can be expressed as a function of \( \frac{y}{x} \). To check this, we can rewrite the equation:
\[ \frac{dy}{dx} = \frac{x^2}{2xy} + \frac{3y^2}{2xy} = \frac{x}{2y} + \frac{3}{2} \frac{y}{x} \]
Now, substituting \( x = \lambda x \) and \( y = \lambda y \):
\[ \frac{dy}{dx} = \frac{\lambda x}{2 \lambda y} + \frac{3}{2} \frac{\lambda y}{\lambda x} = \frac{x}{2y} + \frac{3}{2} \frac{y}{x} \]
This shows that the equation is homogeneous since the terms are invariant under the substitution.

### Step 3: Solve the Differential Equation
To solve the homogeneous equation, we use the substitution \( y = tx \) where \( t = \frac{y}{x} \). Then \( dy = t \, dx + x \, dt \).

Substituting \( y \) and \( dy \) into the differential equation:
\[ t \, dx + x \, dt = \frac{x^2 + 3(tx)^2}{2x(tx)} \, dx \]
This simplifies to:
\[ t + x \frac{dt}{dx} = \frac{x^2 + 3t^2 x^2}{2tx^2} \]
Dividing through by \( x^2 \):
\[ t + x \frac{dt}{dx} = \frac{1 + 3t^2}{2t} \]
Rearranging gives:
\[ x \frac{dt}{dx} = \frac{1 + 3t^2}{2t} - t \]
Combining the terms on the right:
\[ x \frac{dt}{dx} = \frac{1 + 3t^2 - 2t^2}{2t} = \frac{1 + t^2}{2t} \]
This can be rewritten as:
\[ \frac{2t}{1 + t^2} dt = \frac{dx}{x} \]

### Step 4: Integrate Both Sides
Integrating both sides:
\[ \int \frac{2t}{1 + t^2} dt = \int \frac{dx}{x} \]
The left side can be integrated using the substitution \( u = 1 + t^2 \) (where \( du = 2t \, dt \)):
\[ \ln(1 + t^2) = \ln|x| + C \]

### Step 5: Substitute Back for \( y \)
Substituting back \( t = \frac{y}{x} \):
\[ \ln(1 + \left(\frac{y}{x}\right)^2) = \ln|x| + C \]
Exponentiating both sides:
\[ 1 + \frac{y^2}{x^2} = Kx \]
where \( K = e^C \).

### Step 6: Final Form of the Solution
Rearranging gives:
\[ x^2 + y^2 = Kx^3 \]

### Final Solution
Thus, the solution to the differential equation is:
\[ x^2 + y^2 = Cx^3 \]
where \( C \) is a constant.

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