Let y=f(x) be the solution of the differentia | vedclass

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(A) Given the differential equation: $y(x+1) dx = x^2 dy$.Separating the variables,we get: $\frac{x+1}{x^2} dx = \frac{dy}{y}$.Integrating both sides:

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(A) Given the differential equation: $y(x+1) dx = x^2 dy$.Separating the variables,we get: $\frac{x+1}{x^2} dx = \frac{dy}{y}$.Integrating both sides: $\int (\frac{1}{x} + \frac{1}{x^2}) dx = \int \frac{dy}{y}$.This gives: $\ln|x| - \frac{1}{x} = \ln|y| + C$.Using the initial condition $y(1)=e$,we substitute $x=1$ and $y=e$: $\ln(1) - \frac{1}{1} = \ln(e) + C$.$0 - 1 = 1 + C$,which implies $C = -2$.Thus,the solution is $\ln|y| = \ln|x| - \frac{1}{x} + 2$.Taking the exponential of both sides: $y = e^{\ln x - \frac{1}{x} + 2} = x \cdot e^{-\frac{1}{x} + 2}$.Now,we evaluate the limit: $\lim _{x \rightarrow 0^{+}} f(x) = \lim _{x \rightarrow 0^{+}} x \cdot e^{-\frac{1}{x} + 2}$.Let $t = \frac{1}{x}$. As $x \rightarrow 0^{+}$,$t \rightarrow \infty$.The limit becomes $\lim _{t \rightarrow \infty} \frac{e^{-t+2}}{t} = \lim _{t \rightarrow \infty} \frac{e^2}{t e^t} = 0$.

Let $y=f(x)$ be the solution of the differential equation $y(x+1) dx - x^2 dy = 0$ with the initial condition $y(1)=e$. Then $\lim _{x \rightarrow 0^{+}} f(x)$ is equal to

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