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(B) Given the differential equation: $\frac{dy}{dx} = 100 - y$ with initial condition $y(0) = 50$. Separating the variables,we get: $\int \frac{dy}{10

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(B) Given the differential equation: $\frac{dy}{dx} = 100 - y$ with initial condition $y(0) = 50$. Separating the variables,we get: $\int \frac{dy}{100 - y} = \int dx$. Integrating both sides: $-\ln|100 - y| = x + C$. At $x = 0$,$y = 50$,so $-\ln|100 - 50| = 0 + C$,which gives $C = -\ln 50$. Substituting $C$ back into the equation: $-\ln(100 - y) = x - \ln 50$. Rearranging gives: $\ln 50 - \ln(100 - y) = x$,or $\ln(\frac{50}{100 - y}) = x$. Exponentiating both sides: $\frac{50}{100 - y} = e^x$,which implies $100 - y = 50e^{-x}$. Thus,the solution is $y = 100 - 50e^{-x}$. As $x 	o \infty$,$y 	o 100$,and at $x = 0$,$y = 50$. The curve is increasing and approaches the horizontal asymptote $y = 100$. This corresponds to the curve in option $B$.

Which one of the following curves represents the solution of the initial value problem $Dy = 100 - y$,where $y(0) = 50$?

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