MCQ
Which one of the following curves represents the solution of the initial value problem
$Dy = 100 - y,$ where $y (0) = 50$
- A
- ✓
- C
- D
✓
Answer
Correct option: B.
b
$\int {\frac{{dy}}{{100 - y}}} = \int {dx} $ $- \ln (100 - y) = x + C$ $\ln (100 - y) = - x + C$ $x = 0, y = 50$ hence $C = \ln 50$ $x = \ln 50 - \ln (100 - y)$ $\ln \frac{{50}}{{100 - y}} = x \,\, \Rightarrow \frac{{50}}{{100 - y}} = e^x$ $\Rightarrow 100 - y = 50e^{-x} $ $\Rightarrow y = 100 - 50e^{-x}$
$\int {\frac{{dy}}{{100 - y}}} = \int {dx} $ $- \ln (100 - y) = x + C$ $\ln (100 - y) = - x + C$ $x = 0, y = 50$ hence $C = \ln 50$ $x = \ln 50 - \ln (100 - y)$ $\ln \frac{{50}}{{100 - y}} = x \,\, \Rightarrow \frac{{50}}{{100 - y}} = e^x$ $\Rightarrow 100 - y = 50e^{-x} $ $\Rightarrow y = 100 - 50e^{-x}$
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