If the solution curve f(x, y)=0 of the differ | vedclass

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(D) Given the differential equation: $(1+\ln x) \frac{dx}{dy} - x \ln x = e^y$. Let $t = x \ln x$. Then $\frac{dt}{dy} = (1 + \ln x) \frac{dx}{dy}$. S

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$e^{2e^2}$

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(D) Given the differential equation: $(1+\ln x) \frac{dx}{dy} - x \ln x = e^y$. Let $t = x \ln x$. Then $\frac{dt}{dy} = (1 + \ln x) \frac{dx}{dy}$. Substituting this into the equation,we get the linear differential equation: $\frac{dt}{dy} - t = e^y$. The integrating factor is $IF = e^{\int -1 dy} = e^{-y}$. Multiplying by the integrating factor: $e^{-y} \frac{dt}{dy} - e^{-y} t = e^y \cdot e^{-y} = 1$. Integrating both sides with respect to $y$: $\int \frac{d}{dy}(t e^{-y}) dy = \int 1 dy$. $t e^{-y} = y + c$. Substituting $t = x \ln x$: $x \ln x e^{-y} = y + c$,or $x \ln x = (y + c) e^y$. Since the curve passes through $(1, 0)$,we have $1 \ln 1 = (0 + c) e^0$,which implies $0 = c$. Thus,the solution is $x \ln x = y e^y$. For the point $(\alpha, 2)$,we have $\alpha \ln \alpha = 2 e^2$. This can be written as $\ln(\alpha^\alpha) = 2 e^2$. Therefore,$\alpha^\alpha = e^{2e^2}$.

If the solution curve $f(x, y)=0$ of the differential equation $(1+\log_e x) \frac{dx}{dy} - x \log_e x = e^y, x > 0$,passes through the points $(1,0)$ and $(\alpha, 2)$,then $\alpha^\alpha$ is equal to

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