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A window thermometer shows 5 C. It is taken into a room where after a while showing 18 C. The temperature change can be described by the equation dy/dx= - 0.30 (y - 18), where y is the temperature after x minutes. Enter what the thermometer shows two minutes after it is submitted.
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I'm not sure if I made any mistakes, but I have this as a simple seperation of variables problem.\[\frac{dy}{dx}=-0.3(y-18)\to\frac{dy}{y-18}=-0.3dx\]\[\int\frac{dy}{y-18}=\int-0.3dx\to\ln(y-18)=-0.3x+C\]using the intiial value y(0)=5 gives\[y-18=Ce^{-0.3x}\to y=Ce^{-0.3x}+18\]\[y(0)=C+18=5\to C=-13\]\[y=-13e^{-0.3x}+18\]plugging in x=2 then gives\[y(2)=-13e^{-0.6}+18\approx10.87\circ\]
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