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

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Aicharya · Accepted Answer

Given differential equation $\begin{aligned} & y=e^{\left(\frac{d y}{d x}+\frac{d^2 y}{d x^2}ight)} \Rightarrow \ln (y)=\frac{d y}{d x}+\frac{d^2 y}{d x^2} \ & \Rightarrow 	ext { degree }=1, 	ext { order }=2 \ & \Rightarrow \alpha=2, \beta=1 \ & 	ext { Now, } \alpha+\alpha^\beta+\alpha^{2 \beta}+\ldots .+\alpha^{2023 \beta} \ & =2+2^1+2^2+2^3+\ldots+2^{2023} \ & \quad=2+\frac{2\left(2^{2023}-1ight)}{2-1}=2^{2024} . \end{aligned}$

If $\alpha$ and $\beta$ are respectively the order and degree of the differential equation…

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