Question

If \( \alpha \) and \( \beta \) are respectively the order and degree of the differential equation \[ y = e^{\frac{d^2y}{dx^2}}, \] then the value of \( \alpha + \alpha^\beta + \alpha^{2\beta} + \dots + \alpha^{2023\beta} \) is:

• A. \( 2025 \) + 2
• B. \( 2024 \) + 1
• C. \( 2024 \)
• D. \( 2024 \) - 1

Hint:

For geometric progressions, use the formula for the sum of a finite series: \( S = a \frac{r^n - 1}{r - 1} \) where \( a \) is the first term, \( r \) is the common ratio, and \( n \) is the number of terms.

Answer 1

The Correct Option is C

We are given the equation: \[ y = e^{\frac{d^2y}{dx^2}} \] Step 1: To find the order and degree of the differential equation: - The equation involves the second derivative of \( y \), so the order is 2. - The degree is the power of the highest order derivative, which is 1 in this case because the second derivative appears in the exponent. Thus, \( \alpha = 2 \) and \( \beta = 1 \). Step 2: Calculate the sum: We need to compute the value of \( \alpha + \alpha^\beta + \alpha^{2\beta} + \dots + \alpha^{2023\beta} \). This is a sum of terms in a geometric progression with first term \( \alpha \) and common ratio \( \alpha^\beta \), so: \[ S = \alpha + \alpha^\beta + \alpha^{2\beta} + \dots + \alpha^{2023\beta} \] Substituting \( \alpha = 2 \) and \( \beta = 1 \), we get: \[ S = 2 + 2^1 + 2^2 + \dots + 2^{2023} \] This is the sum of the first 2024 terms of a geometric progression with the first term 2 and common ratio 2, which equals: \[ S = 2^{2024} - 1 \] Thus, the correct answer is \( 2024 \). % Final Answer The value is \( 2024 \).