## Related questions with answers

Hospital patients are often given glucose (blood sugar) through a tube connected to a bottle suspended over their beds. Suppose that this “drip” supplies glucose at the rate of 25 mg per minute, and each minute 10% of the accumulated glucose is consumed by the body. Then the amount y(t) of glucose (in excess of the normal level) in the body after t minutes satisfies $y^{\prime}=25-0.1 y$ (Do you see why?) $y(0) = 0$ (zero excess glucose at t = 0) Solve this differential equation and initial condition.

Solution

Verifieda)\ Then the amount y(t) of glucose (in excess of the normal level) in the body after t minutes satisfies$\boxed{y'=25-0.1y}$.initial condition $y(0)=0$. we need to solve the differential equation

$\begin{align*} y'&=25-0.1y\\ \dfrac{dy}{dt}&=25-0.1y\\ \dfrac{dy}{(25-0.1y)}&=1 .dt\\ \int \dfrac{dy}{(25-0.1y)}&=\int 1.dt \ \ \ \text{(Integrate on both side of equation)}\\ \end{align*}$

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