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Chapter 2: Problem 15
\((\cosh 6 t+5 \sinh 4 t) d t+20 \sinh y d y=0\)
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\(2 \ln t d t-\ln \left(4 y^{2}\right) d y=0\)
\(\left(e^{t}+1\right) d y / d t+e^{t} y=t, y(0)=-1\)
Graph the solution of \(y^{\prime}=\sin (x y)\) subject to the initial condition \(y(0)=i\) on the interval \([0,7]\) for \(i=0.5,1.0,1.5,2.0\), and \(2.5\). In each case, approximate the value of the solution if \(x=0.5\).
Suppose that a drug is added to the body at a rate \(r(t)\), and let \(y(t)\) represent the concentration of the drug in the bloodstream at time \(t\) hours. In addition, suppose that the drug is used by the body at the rate \(k y\), where \(k\) is a positive constant. Then, the net rate of change in \(y(t)\) is given by the equation \(d y / d t=r(t)-k y\). If at \(t=0\), there is no drug in the body, we determine \(y(t)\) by solving the initial value problem $$ d y / d t=r(t)-k y, \quad y(0)=0 . $$ (a) Suppose that \(r(t)=r\), where \(r\) is a positive constant. In this case, the drug is added at a constant rate. Sketch the phase line for \(d y / d t=r-k y\). Solve the IVP. Determine \(\lim _{t \rightarrow \infty} y(t)\). How does this limit correspond to the phase line? (b) Suppose that \(r(t)=1+\sin t\) and \(k=\) 1. In this case, the drug is added at a periodic rate. Solve the IVP. Determine \(\lim _{t \rightarrow \infty} y(t)\), if it exists. Describe what happens to the drug concentration over time. (c) Suppose that \(r(t)=e^{-t}\) and \(k=1\). In this case, the rate at which the drug is added decreases over time. Solve the IVP. Determine \(\lim _{t \rightarrow \infty} y(t)\), if it exists. Describe what happens to the drug concentration over time.
Equations of the form \(y=t f\left(y^{\prime}\right)+g\left(y^{\prime}\right)\) are called Lagrange equations. These equations are solved by making the substitution \(p=y^{\prime}(t)\). (a) Differentiate \(y=t f\left(y^{\prime}\right)+g\left(y^{\prime}\right)\) with respect to \(t\) to obtain $$ y^{\prime}=t f^{\prime}\left(y^{\prime}\right) y^{\prime \prime}+f\left(y^{\prime}\right)+g^{\prime}\left(y^{\prime}\right) y^{\prime \prime} . $$ (b) Substitute \(p\) into the equation to obtain $$ \begin{aligned} p &=t f^{\prime}(p) \frac{d p}{d t}+f(p)+g^{\prime}(p) \frac{d p}{d t} \\ &=f(p)+\left(t f^{\prime}(p)+g^{\prime}(p)\right) \frac{d p}{d t} \end{aligned} $$ (c) Solve this equation for \(d t / d p\) to obtain the linear equation $$ \frac{d t}{d p}=\frac{t f^{\prime}(p)+g^{\prime}(p)}{p-f(p)} $$ which is equivalent to $$ \frac{d t}{d p}+\frac{f^{\prime}(p)}{f(p)-p} t=\frac{g^{\prime}(p)}{p-f(p)} . $$ This first order linear equation can be solved for \(t\) in terms of \(p\). Then, \(t(p)\) can be used with \(y=t f(p)+g(p)\) to obtain an equation for \(y\).
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