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(C) Given the differential equation: $y = e^{\left(\frac{dy}{dx} + \frac{d^2y}{dx^2}\right)}$Taking the natural logarithm on both sides: $\ln(y) = \fr

Vedclass · Accepted Answer

$2^{2024}$

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(C) Given the differential equation: $y = e^{\left(\frac{dy}{dx} + \frac{d^2y}{dx^2}\right)}$Taking the natural logarithm on both sides: $\ln(y) = \frac{dy}{dx} + \frac{d^2y}{dx^2}$The highest order derivative present is $\frac{d^2y}{dx^2}$,so the order $\alpha = 2$.The power of the highest order derivative is $1$,so the degree $\beta = 1$.We need to find the sum $S = \alpha + \alpha^\beta + \alpha^{2\beta} + \ldots + \alpha^{2023\beta}$.Substituting $\alpha = 2$ and $\beta = 1$: $S = 2 + 2^1 + 2^2 + 2^3 + \ldots + 2^{2023}$.This is a geometric progression with $2024$ terms,where the first term $a = 2$,common ratio $r = 2$,and number of terms $n = 2024$.The sum is $S = 2 + (2^1 + 2^2 + \ldots + 2^{2023}) = 2 + \frac{2(2^{2023} - 1)}{2 - 1} = 2 + 2^{2024} - 2 = 2^{2024}$.

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

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