The number of solutions of y' = frac{y + 1}{x | vedclass

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(A) Given the differential equation $\frac{dy}{dx} = \frac{y + 1}{x - 1}$.Separating the variables,we get $\frac{dy}{y + 1} = \frac{dx}{x - 1}$.Integr

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(A) Given the differential equation $\frac{dy}{dx} = \frac{y + 1}{x - 1}$.Separating the variables,we get $\frac{dy}{y + 1} = \frac{dx}{x - 1}$.Integrating both sides,$\int \frac{dy}{y + 1} = \int \frac{dx}{x - 1}$.This gives $\ln|y + 1| = \ln|x - 1| + C$,which simplifies to $y + 1 = k(x - 1)$ where $k$ is a constant.We are given the initial condition $y(1) = 2$.Substituting $x = 1$ into the equation,we get $y + 1 = k(1 - 1) = 0$,which implies $y = -1$.However,the condition specifies $y(1) = 2$,which contradicts $y = -1$.Since the differential equation is undefined at $x = 1$ and the initial condition leads to a contradiction,there is no solution.

The number of solutions of $y' = \frac{y + 1}{x - 1}, y(1) = 2$ is

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