The substituion `y=z^(alpha)` transforms the differential equation `(x^(2)y^(2)-1)dy+2xy^(3)dx=0` into a homogeneous differential equation for

Options

A. `alpha=-1`
B. 0
C. `alpha=1`
D. No value of `alpha`

Answer

To solve the given differential equation \((x^2 y^2 - 1) dy + 2xy^3 dx = 0\) using the substitution \(y = z^\alpha\), we will follow these steps:

### Step 1: Rewrite the Differential Equation
Start with the given differential equation:
\[ (x^2 y^2 - 1) dy + 2xy^3 dx = 0 \]

### Step 2: Express \(dy\) in Terms of \(dx\)
Rearranging gives:
\[ dy = -\frac{2xy^3}{x^2 y^2 - 1} dx \]

### Step 3: Substitute \(y = z^\alpha\)
Now, substitute \(y = z^\alpha\) into the equation. We need to find \(dy\):
\[ dy = \alpha z^{\alpha - 1} dz \]
Substituting \(y = z^\alpha\) into the differential equation gives:
\[ (x^2 (z^\alpha)^2 - 1) \alpha z^{\alpha - 1} dz + 2x(z^\alpha)^3 dx = 0 \]
This simplifies to:
\[ (x^2 z^{2\alpha} - 1) \alpha z^{\alpha - 1} dz + 2xz^{3\alpha} dx = 0 \]

### Step 4: Rearranging the Equation
Rearranging the equation, we have:
\[ \alpha (x^2 z^{2\alpha} - 1) z^{\alpha - 1} dz + 2xz^{3\alpha} dx = 0 \]
This can be expressed as:
\[ \alpha (x^2 z^{2\alpha} - 1) dz = -2xz^{3\alpha} dx \]

### Step 5: Forming a Homogeneous Equation
Dividing through by \(z^{3\alpha}\) gives:
\[ \alpha \frac{x^2 z^{2\alpha}}{z^{3\alpha}} - \frac{\alpha}{z^{3\alpha}} = -2x \frac{dx}{dz} \]
This simplifies to:
\[ \alpha x^2 z^{-1} - \alpha z^{-3\alpha} = -2x \frac{dx}{dz} \]
Now, we can express this in a homogeneous form.

### Step 6: Identify Homogeneous Conditions
For the differential equation to be homogeneous, we need to check the condition:
If we substitute \(x\) by \(\lambda x\) and \(z\) by \(\lambda z\), the equation should remain unchanged. This leads us to:
\[ \alpha \lambda^2 z^{-1} - \alpha \lambda^{-3\alpha} = -2\lambda x \frac{dx}{dz} \]

### Step 7: Equate Powers of \(\lambda\)
From the above equation, we need the powers of \(\lambda\) to match:
\[ 0 = 2 + 2\alpha \]
Solving this gives:
\[ 2\alpha = -2 \implies \alpha = -1 \]

### Conclusion
Thus, the substitution \(y = z^{-1}\) transforms the given differential equation into a homogeneous differential equation.