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 find the value of \( \alpha \) such that \( y = e^{(\alpha + 1)x} \) is a solution of the differential equation

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

we will follow these steps:

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

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

Differentiating with respect to \( x \):
\[ \frac{dy}{dx} = (\alpha + 1)e^{(\alpha + 1)x} \]

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

Now, differentiate \( \frac{dy}{dx} \):
\[ \frac{d^2y}{dx^2} = \frac{d}{dx} \left((\alpha + 1)e^{(\alpha + 1)x}\right) = (\alpha + 1)^2 e^{(\alpha + 1)x} \]

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

Substituting into the differential equation:
\[ (\alpha + 1)^2 e^{(\alpha + 1)x} - 4(\alpha + 1)e^{(\alpha + 1)x} + 4e^{(\alpha + 1)x} = 0 \]

### Step 4: Factor out \( e^{(\alpha + 1)x} \)

Since \( e^{(\alpha + 1)x} \neq 0 \), we can factor it out:
\[ e^{(\alpha + 1)x} \left((\alpha + 1)^2 - 4(\alpha + 1) + 4\right) = 0 \]

### Step 5: Simplify the equation inside the brackets

Now, we simplify the expression:
\[ (\alpha + 1)^2 - 4(\alpha + 1) + 4 = 0 \]
Expanding \( (\alpha + 1)^2 \):
\[ \alpha^2 + 2\alpha + 1 - 4\alpha - 4 + 4 = 0 \]
This simplifies to:
\[ \alpha^2 - 2\alpha + 1 = 0 \]

### Step 6: Factor the quadratic equation

Factoring gives:
\[ (\alpha - 1)^2 = 0 \]

### Step 7: Solve for \( \alpha \)

Setting the factor equal to zero:
\[ \alpha - 1 = 0 \implies \alpha = 1 \]

Thus, the value of \( \alpha \) is:
\[ \boxed{1} \]