If `y=e ^((alpha +1)x)` be solution of differential equation `(d ^(2)y)/(dx ^(2)) -4 (dy )/(dx) +4y=0,` then `alpha` is:

Options

A. 0
B. 1
C. `-1`
D. 2

Answer

To solve the problem, we need to determine the value of `alpha` such that the function \( y = e^{(\alpha + 1)x} \) satisfies the differential equation:

\[ \frac{d^2y}{dx^2} - 4 \frac{dy}{dx} + 4y = 0 \]

### Step 1: Find the first derivative \( \frac{dy}{dx} \)

Given:
\[ y = e^{(\alpha + 1)x} \]

Using the chain rule, we differentiate \( y \):
\[ \frac{dy}{dx} = (\alpha + 1)e^{(\alpha + 1)x} \]

### Step 2: Find the second derivative \( \frac{d^2y}{dx^2} \)

Now we differentiate \( \frac{dy}{dx} \):
\[ \frac{d^2y}{dx^2} = \frac{d}{dx} \left( (\alpha + 1)e^{(\alpha + 1)x} \right) \]
Using the chain rule again:
\[ \frac{d^2y}{dx^2} = (\alpha + 1)^2 e^{(\alpha + 1)x} \]

### Step 3: Substitute \( y \), \( \frac{dy}{dx} \), and \( \frac{d^2y}{dx^2} \) into the differential equation

Now we substitute these derivatives back into the differential equation:
\[ (\alpha + 1)^2 e^{(\alpha + 1)x} - 4(\alpha + 1)e^{(\alpha + 1)x} + 4e^{(\alpha + 1)x} = 0 \]

Factoring out \( e^{(\alpha + 1)x} \) (which is never zero):
\[ e^{(\alpha + 1)x} \left( (\alpha + 1)^2 - 4(\alpha + 1) + 4 \right) = 0 \]

### Step 4: Solve the quadratic equation

Now we solve the quadratic equation:
\[ (\alpha + 1)^2 - 4(\alpha + 1) + 4 = 0 \]

Let \( z = \alpha + 1 \):
\[ z^2 - 4z + 4 = 0 \]

This simplifies to:
\[ (z - 2)^2 = 0 \]

Thus:
\[ z - 2 = 0 \implies z = 2 \]

Substituting back for \( \alpha \):
\[ \alpha + 1 = 2 \implies \alpha = 1 \]

### Conclusion

The value of \( \alpha \) is:
\[ \boxed{1} \]