If α and β are respectively the order and degree of the differential equation y=edydx+d2ydx2, then the value of α+αβ+α2β+…+α2023β=

y=edydx+d2ydx2logey=dydx+d2ydx2d2ydx2+dydx+logey=0Order=2 and degree=1=α+αβ+α2β+…+α2023βα=2 and β=1=2+2+22+23+···+22023

If α and β are respectively the order and degree of the differential equation y=edydx+d2ydx2, then the value of α+αβ+α2β+…+α2023β=

y=edydx+d2ydx2logey=dydx+d2ydx2d2ydx2+dydx+logey=0Order=2 and degree=1=α+αβ+α2β+…+α2023βα=2 and β=1=2+2+22+23+···+22023

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If $α and β$ are respectively the order and degree of the differential equation $y = e^{(\frac{d y}{d x} + \frac{d^{2} y}{d x^{2}})}$ , then the value of $α + α^{β} + α^{2 β} + \dots + α^{2023 β} =$

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$2^{2024} - 1$

$2^{2024}$

$2^{2024} + 1$

$2^{2025} + 2$

answer is C.

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Detailed Solution

$y = e^{(\frac{d y}{d x} + \frac{d^{2} y}{d x^{2}})}$

$\log_{e} y = (\frac{d y}{d x} + \frac{d^{2} y}{d x^{2}})$

$\frac{d^{2} y}{d x^{2}} + \frac{d y}{d x} + \log_{e} y = 0$

Order=2 and degree=1

$= α + α^{β} + α^{2 β} + \dots + α^{2023 β}$

$α = 2$ and $β = 1$

$= 2 + 2 + 2^{2} + 2^{3} + \cdot \cdot \cdot + 2^{2023}$

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If α and β are respectively the order and degree of the differential equation y=edydx+d2ydx2, then the value of α+αβ+α2β+…+α2023β=

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If $α and β$ are respectively the order and degree of the differential equation $y = e^{(\frac{d y}{d x} + \frac{d^{2} y}{d x^{2}})}$ , then the value of $α + α^{β} + α^{2 β} + \dots + α^{2023 β} =$