1.3: Applications

In this section we will look at some simple applications which are modeled with first order differential equations. We will begin with simple exponential models of growth and decay.

Growth and Decay

Some of the simplest models are those involving growth or decay. For example, a population model can be obtained under simple assumptions. Let \(P(t)\) be the population at time \(t\). We want to find an expression for the rate of change of the population, \(\dfrac{d P}{d t}\). Assuming that there is no migration of population, the only way the population can change is by adding or subtracting individuals in the population. The equation would take the form

\[\dfrac{d P}{d t}=\text { Rate In }-\text { Rate Out. } \nonumber \]

The Rate In could be due to the number of births per unit time and the Rate Out by the number of deaths per unit time. The simplest forms for these rates would be given by

\[\text { Rate In }=b P \text { and the Rate Out }=m P . \nonumber \]

Here we have denoted the birth rate as \(b\) and the mortality rate as \(m\). This gives the total rate of change of population as

\[\dfrac{d P}{d t}=b P-m P \equiv k P \label{1.27} \]

Equation \(\PageIndex{1}\) is a separable equation. The separation follows as we have seen earlier in the chapter. Rearranging the equation, its differential form is

\[\dfrac{d P}{P}=k d t \nonumber \]

Integrating, we have

\[ \begin{aligned} &\int \dfrac{d P}{P}=\int k d t \\[4pt] &\ln |P|=k t+C \end{aligned} \label{1.28} \]

Next, we solve for \(P(t)\) through exponentiation, Integrating, we have

\[ \begin{aligned} |P(t)| &=e^{k t+C} \\[4pt] P(t) &=\pm e^{k t+C} \\[4pt] &=\pm e^{C} e^{k t} \\[4pt] &=A e^{k t} . \end{aligned}\label{1.29} \]

More generally, the initial value problem \(d P / d t=k P, P\left(t_{0}\right)=P_{0}\) has the solution \(P(t) = P_0 e^{k(t-t_0)}\).

Here we renamed the arbitrary constant, \(\pm e^{C}\), as \(A \).

If the population at \(t=0\) is \(P_{0}\), i.e., \(P(0)=P_{0}\), then the solution gives \(P(0)=A e^{0}=A=P_{0} \). So, the solution of the initial value problem is

\[P(t)=P_{0} e^{k t} \nonumber \]

Malthusian population growth.

Equation \(\PageIndex{1}\) the familiar exponential model of population growth:

\[\dfrac{d P}{d t}=k P \nonumber \]

This is easily solved and one obtains exponential growth \((k>0)\) or decay \((k<0)\).

This Malthusian growth model has been named after Thomas Robert Malthus \((1766-1834)\), a clergyman who used this model to warn of the impending doom of the human race if its reproductive practices continued.

Example \(\PageIndex{1}\)

Consider a bacteria population of weight \(20 \mathrm{~g}\). If the population doubles every 20 minutes, then what is the population after 30 minutes? [Note: It is easier to weigh this population than to count it.]

Solution

One looks at the given information before trying to answer the question. First, we have the initial condition \(P_{0}=20 \mathrm{~g}\). Since the population doubles every 20 minutes, then \(P(20)=2 P_{0}=40\). Here we have take the time units as minutes. We are then asked to find \(P(30)\).

We do not need to solve the differential equation. We will assume a simple growth model. Using the general solution, \(P(t)=20 e^{k t}\), we have

\[P(20)=20 e^{20 k}=40 \nonumber \]

or

\[e^{20 k}=2 \nonumber \]

We can solve this for \(k\),

\[20 k=\ln 2, \quad \Rightarrow k=\dfrac{\ln 2}{20} \approx 0.035 \nonumber \]

This gives an approximate solution, \(P(t) \approx 20 e^{.035 t} \). Now we can answer the original question. Namely, \(P(30) \approx 57 \).

Of course, we could get an exact solution. With some simple manipulations, we have

\[ \begin{aligned} P(t) &=20 e^{k t} \\[4pt] &=20 e^{\left(\dfrac{\ln 2}{20}\right) t} \\[4pt] &=20\left(e^{\ln 2}\right)^{\dfrac{t}{20}} \\[4pt] &=20\left(2^{\dfrac{t}{20}}\right) \end{aligned} \label{1.30} \]

This answer takes the general form for population doubling, \(P(t)=\) \(P_{0} 2^{\dfrac{t}{\tau}}\), where \(\tau\) is the doubling rate.

Another standard growth-decay problem is radioactive decay. Certain isotopes are unstable and the nucleus breaks apart, leading to nuclear decay. The products of the decay may also be unstable and undergo further nuclear decay. As an example, Uranium-238 (U-238) decays into Thorium-234 (Th-Radioactive decay problems. 234). Thorium-234 is unstable and decays into Protactinium (Pa-234). This in turn decays in many steps until lead (Pb-206) is produced as shown in Table \(\PageIndex{1}\). This lead isotope is stable and the decay process stops. While this is one form of radioactive decay, there are other types. For example, Radon 222 (Rn-222) gives up an alpha particle (helium nucleus) leaving Polonium (Po-218).

Table \(\PageIndex{1}\): U-238 decay chain.

Isotope	Half-life
\(U^{238}\)	\(4.468 x 10^{9}\) years
\(T h^{234}\)	\(24.1\) days
\(P a^{234 m}\)	\(1.17\) minutes
\(U^{234}\)	\(2.47 x 10^{5}\) years
\(T h^{230}\)	\(8.0 x 10^{4}\) years
\(R a^{226}\)	1602 years
\(R n^{222}\)	\(3.823\) days
\(P o^{218}\)	\(3.05\) minutes
\(P b^{214}\)	\(26.8\) minutes
\(B i^{214}\)	\(19.7\) minutes
\(P o^{214}\)	164 microsec
\(P b^{210}\)	21 years
\(B i^{210}\)	\(5.01\) days
\(P o^{210}\)	\(138.4\) days
\(P b^{206}\)	stable

Given a certain amount of radioactive material, it does not all decay at one time. A measure of the tendency of a nucleus to decay is called the half-life. This is the time it takes for half of the material to decay. This is similar to the last example and can be understood using a simple example.

Example \(\PageIndex{2}\)

If \(150.0 \mathrm{~g}\) of Thorium-234 decays to \(137.6 \mathrm{~g}\) of Thorium234 in three days, what is its half-life?

Solution

This is another simple decay process. If \(Q(t)\) represents the quantity of unstable material, then \(Q(t)\) satisfies the rate equation

\[\dfrac{d Q}{d t}=k Q \nonumber \]

with \(k<0 \). The solution of the initial value problem, as we have seen, is \(Q(t)=Q_{0} e^{k t}\).

Now, let the half-life be given by \(\tau\). Then, \(Q(\tau)=\dfrac{1}{2} Q_{0}\). Inserting this fact into the solution, we have

\[ \begin{aligned} Q(\tau) &=Q_{0} e^{k \tau} \\[4pt] \dfrac{1}{2} Q_{0} &=Q_{0} e^{k \tau} \\[4pt] \dfrac{1}{2} &=e^{k \tau} \end{aligned} \label{1.31} \]

Noting that \(Q(t)=Q_{0}\left(e^{k}\right)^{t}\), we solve Equation \(\PageIndex{5}\) for

\[e^{k}=2^{-1 / \tau} \nonumber \]

Then, the solution can be written in the general form

\[Q(t)=Q_{0} 2^{-\dfrac{t}{\tau}} \nonumber \]

Note that the decay constant is \(k=-\dfrac{\ln 2}{\tau}<0\).

Returning to the problem, we are given

\[Q(3)=1502^{-\dfrac{3}{\tau}}=137.6 \nonumber \]

Solving to \(\tau\),

\[ \begin{aligned} 2^{-\dfrac{3}{\tau}} &=\dfrac{136.7}{150} \\[4pt] -3 \ln 2 &=\ln .9173 \tau \\[4pt] \tau &=-\dfrac{3 \ln 2}{\ln .9173}=24.09 \end{aligned} \nonumber \]

Therefore, the half-life is about \(24.1\) days.

Newton’s Law of Cooling

If you take your hot cup of tea, and let it sit in a cold room, the tea will cool off and reach room temperature after a period of time. The law of cooling is attributed to Isaac Newton (1642-1727) who was probably the first to state results on how bodies cool. \({ }^{1}\) The main idea is that a body attemperature \(T(t)\) is initially at temperature \(T(0)=T_{0} \). It is placed in an environment at an ambient temperature of \(T_{a}\). A simple model is given that the rate of change of the temperature of the body is proportional to the difference between the body temperature and its surroundings. Thus, we have

\[\dfrac{dT}{dt} \propto T -T_a \nonumber \]

The proportionality is removed by introducing a cooling constant.

\[\dfrac{d T}{d t}=-k\left(T-T_{a}\right) \label{1.33} \]

where \(k>0\).

Note 1

Newton’s 1701 Law of Cooling is an approximation to how bodies cool for small temperature differences \(\left(T-T_{a} \ll T\right)\) and does not take into account all of the cooling processes. One account is given by C. T. O’Sullivan, Am. J. Phys (1990) p 956-960.

This differential equation can be solved by noting that the equation can be written in the form

\[\dfrac{d}{d t}\left(T-T_{a}\right)=-k\left(T-T_{a}\right) \nonumber \]

This is now of the form of exponential decay of the function \(T(t)-T_{a}\). The solution is easily found as

\[T(t)-T_{a}=\left(T_{0}-T_{a}\right) e^{-k t} \nonumber \]

Or

\[T(t)=T_{a}+\left(T_{0}-T_{a}\right) e^{-k t} \nonumber \]

Example \(\PageIndex{3}\)

A cup of tea at \(90^{\circ} \mathrm{C}\) cools to \(85^{\circ} \mathrm{C}\) in ten minutes. If the room temperature is \(22^{\circ} \mathrm{C}\), what is its temperature after 30 minutes?

Solution

Using the general solution with \(T_{0}=90^{\circ} \mathrm{C}\),

\[T(t)=22+(90-22) e^{-k}=22+68 e^{-k t} \nonumber \]

we then find \(k\) using the given information, \(T(10)=85^{\circ} \mathrm{C}\). We have

\[ \begin{aligned} 85 &=T(10) \\[4pt] &=22+68 e^{-10 k} \\[4pt] 63 &=68 e^{-10 k} \\[4pt] e^{-10 k} &=\dfrac{63}{68} \approx 0.926 \\[4pt] -10 k &=\ln 0.926 \\[4pt] k &=-\dfrac{\ln 0.926}{10}=0.00764 \end{aligned}\label{1.34} \]

This gives the equation for this model as

\[[T(t)=22+68 e^{-0.00764 t} \nonumber\]

Now we can answer the question. What is \(T(30) ?\)

\[T(30)=22+68 e^{-0.00764(30)}=76^{\circ} \mathrm{C} \nonumber \]

Terminal Velocity

Now let’s return to free fall. What if there is air resistance? We first need to model the air resistance. As an object falls faster and faster, the drag force becomes greater. So, this resistive force is a function of the velocity. There are a couple of standard models that people use to test this. The idea is to write \(F=m a\) in the form

\[m \ddot{y}=-m g+f(v) \nonumber \]

where \(f(v)\) gives the resistive force and \(m g\) is the weight. Recall that this applies to free fall near the Earth’s surface. Also, for it to be resistive, \(f(v)\) should oppose the motion. If the body is falling, then \(f(v)\) should be positive. If it is rising, then \(f(v)\) would have to be negative to indicate the opposition to the motion.

One common determination derives from the drag force on an object moving through a fluid. This force is given by

\[f(v)=\dfrac{1}{2} C A \rho v^{2} \nonumber \]

where \(C\) is the drag coefficient, \(A\) is the cross sectional area and \(\rho\) is the fluid density. For laminar flow the drag coefficient is constant.

Unless you are into aerodynamics, you do not need to get into the details of the constants. So, it is best to absorb all of the constants into one to simplify the computation. So, we will write \(f(v)=b v^{2}\). The differential equation including drag can then be rewritten as

\[\dot{v}=k v^{2}-g \nonumber \]

where \(k=b / m\). Note that this is a first order equation for \(v(t)\). It is separable too! Formally, we can separate the variables and integrate over time to obtain

\[t+K=\int^{v} \dfrac{d z}{k z^{2}-g} \nonumber \]

This is the first use of Partial Fraction Decomposition. We will explore this method further in the section on Laplace Transforms.

(Note: We used an integration constant of \(K\) since \(C\) is the drag coefficient in this problem.) If we can do the integral, then we have a solution for v. In fact, we can do this integral. You need to recall another common method of integration, which we have not reviewed yet. Do you remember Partial Fraction Decomposition? It involves factoring the denominator in the integral. In the simplest case there are two linear factors in the denominator and the integral is rewritten:

\[\int \dfrac{d x}{(x-a)(x-b)}=\dfrac{1}{b-a} \int\left[\dfrac{1}{x-a}-\dfrac{1}{x-b}\right] d x \nonumber \]

The new integral now has two terms which can be readily integrated.

In order to factor the denominator in the current problem, we first have to rewrite the constants. We let \(\alpha^{2}=g / k\) and write the integrand as

\[\dfrac{1}{k z^{2}-g}=\dfrac{1}{k} \dfrac{1}{z^{2}-\alpha^{2}} \nonumber \]

Now we use a partial fraction decomposition to obtain

\[\dfrac{1}{k z^{2}-g}=\dfrac{1}{2 \alpha k}\left[\dfrac{1}{z-\alpha}-\dfrac{1}{z+\alpha}\right] \nonumber \]

Now, the integrand can be easily integrated giving

\[t+K=\dfrac{1}{2 \alpha k} \ln \left|\dfrac{v-\alpha}{v+\alpha}\right| \nonumber \]

Solving for \(v\), we have

\[v(t)=\dfrac{1-B e^{2 \alpha k t}}{1+B e^{2 \alpha k t}} \alpha \nonumber \]

where \(B \equiv e^{K} . B\) can be determined using the initial velocity.

There are other forms for the solution in terms of a tanh function, which the reader can determine as an exercise. One important conclusion is that for large times, the ratio in the solution approaches \(-1 \). Thus, \(v \rightarrow-\alpha=\) \(-\sqrt{\dfrac{g}{k}}\) as \(t \rightarrow \infty\). This means that the falling object will reach a constant terminal velocity.

As a simple computation, we can determine the terminal velocity. We will take an \(80 \mathrm{~kg}\) skydiver with a cross sectional area of about \(0.093 \mathrm{~m}^{2}\). (The skydiver is falling head first.) Assume that the air density is a constant \(1.2 \mathrm{~kg} / \mathrm{m}^{3}\) and the drag coefficient is \(C=2.0\). We first note that

\[v_{\text {terminal }}=-\sqrt{\dfrac{g}{k}}=-\sqrt{\dfrac{2 m g}{C A \rho}} \nonumber \]

So,

\[v_{\text {terminal }}=-\sqrt{\dfrac{2(70)(9.8)}{(2.0)(0.093)(1.2)}}=-78 \mathrm{~m} / \mathrm{s} \nonumber \]

This is about \(175 \mathrm{mph}\), which is slightly higher than the actual terminal velocity of a sky diver with arms and feet fully extended. One would need a more accurate determination of \(C\) and \(A\) for a more realistic answer. Also, the air density varies along the way.

Mixture Problems

Mixture problems often occur in a first course on differential equations as examples of first order differential equations. In such problems we consider a tank of brine, water containing a specific amount of salt with pure water entering and the mixture leaving, or the flow of a pollutant into, or out of, a lake. The goal is to predict the amount of salt, or pollutant, at some later time.

In general one has a rate of flow of some concentration of mixture entering a region and a mixture leaving the region. The goal is to determine how much stuff is in the region at a given time. This is governed by the equation

Theorem \(\PageIndex{1}\)

Rate of change of substance \(=\) Rate In \(-\) Rate Out.

The rates are not often given. One is generally given information about the concentration and flow rates in and out of the system. If one pays attention to the dimension and sketches the situation, then one can write out this rate equation as a first order differential equation. We consider a simple example.

Example \(\PageIndex{4}\): Single Tank Problem

A 50 gallon tank of pure water has a brine mixture with concentration of 2 pounds per gallon entering at the rate of 5 gallons per minute. [See Figure \(\PageIndex{1}\).] At the same time the well-mixed contents drain out at the rate of 5 gallons per minute. Find the amount of salt in the tank at time \(t \). In all such problems one assumes that the solution is well mixed at each instant of time.

Figure \(\PageIndex{1}\): A typical mixing problem

Solution

Let \(x(t)\) be the amount of salt at time \(t\). Then the rate at which the salt in the tank increases is due to the amount of salt entering the tank less that leaving the tank. To figure out these rates, one notes that \(d x / d t\) has units of pounds per minute. The amount of salt entering per minute is given by the product of the entering concentration times the rate at which the brine enters. This gives the correct units:

\[\left(2 \dfrac{\text { pounds }}{\text { gal }}\right)\left(5 \dfrac{\text { gal }}{\text { min }}\right)=10 \dfrac{\text { pounds }}{\text { min }} . \nonumber \]

Similarly, one can determine the rate out as

\[\left(\dfrac{x \text { pounds }}{50 \text { gal }}\right)\left(5 \dfrac{\text { gal }}{\text { min }}\right)=\dfrac{x}{10} \dfrac{\text { pounds }}{\text { min }} . \nonumber \]

Thus, we have

\[\dfrac{d x}{d t}=10-\dfrac{x}{10} \nonumber \]

This equation is solved using the methods for linear first order equations. The integrating factor is \(\mu=e^{x / 10}\), leading to the general solution

\[x(t)=100+A e^{-t / 10} \nonumber \]

Using the initial condition, one finds the particular solution

\[x(t)=100\left(1-e^{-t / 10}\right) \nonumber \]

Often one is interested in the long time behavior of a system. In this case we have that \(\lim _{t \rightarrow \infty} x(t)=100 \mathrm{lb}\). This makes sense because 2 pounds per galloon enter during this time to eventually leave the entire 50 gallons with this concentration. Thus,

\[50 \mathrm{gal} \times 2 \dfrac{\mathrm{lb}}{50 \mathrm{gal}}=100 \mathrm{lb} \nonumber \]

Orthogonal Trajectories of Curves

There are many problems from geometry which have lead to the study of differential equations. One such problem is the construction of orthogonal trajectories. Give a a family of curves, \(y_{1}(x ; a)\), we seek another family of curves \(y_{2}(x ; c)\) such that the second family of curves are perpendicular the to given family. This means that the tangents of two intersecting curves at the point of intersection are perpendicular to each other. The slopes of the tangent lines are given by the derivatives \(y_{1}^{\prime}(x)\) and \(y_{2}^{\prime}(x)\). We recall from elementary geometry that the slopes of two perpendicular lines are related by

\[y_{2}^{\prime}(x)=-\dfrac{1}{y_{1}^{\prime}(x)} \nonumber \]

Example \(\PageIndex{5}\)

Find a family of orthogonal trajectories to the family of parabolae \(y_{1}(x ; a)=a x^{2}\).

Solution

We note that the new collection of curves has to satisfy the equation

\[y_{2}^{\prime}(x)=-\dfrac{1}{y_{1}^{\prime}(x)}=-\dfrac{1}{2 a x} \nonumber \]

Before solving for \(y_{2}(x)\), we need to eliminate the parameter \(a\). From the give function, we have that \(a=\dfrac{y}{x^{2}}\). Inserting this into the equation for \(y_{2}^{\prime}\), we have

\[y^{\prime}(x)=-\dfrac{1}{2 a x}=-\dfrac{x}{2 y}\nonumber \]

Thus, to find \(y_{2}(x)\), we have to solve the differential equation

\[2 y y^{\prime}+x=0\nonumber \]

Noting that \(\left(y^{2}\right)^{\prime}=2 y y^{\prime}\) and \(\left(\dfrac{1}{2} x^{2}\right)^{\prime}=x_{\prime \prime}\) this (exact) equation can be Written as

\[\dfrac{d}{d x}\left(y^{2}+\dfrac{1}{2} x^{2}\right)=0\nonumber \]

Integrating, we find the family of solutions,

\[y^{2}+\dfrac{1}{2} x^{2}=k \nonumber \]

In Figure \(\PageIndex{2}\) we plot both families of orthogonal curves.

Figure \(\PageIndex{2}\): Plot of orthogonal families of curves, \(y=a x^{2}\) and \(y^{2}+\dfrac{1}{2} x^{2}=k\).

Pursuit Curves*

Another application that is interesting is to find the path that a body traces out as it moves towards a fixed point or another moving body. Such curses are know as pursuit curves. These could model aircraft or submarines following targets, or predators following prey. We demonstrate this with an example.

Example \(\PageIndex{6}\)

A hawk at point \((x, y)\) sees a sparrow traveling at speed \(v\) along a straight line. The hawk flies towards the sparrow at constant speed \(w\) but always in a direction along line of sight between their positions. If the hawk starts out at the point \((a, 0)\) at \(t=0\), when the sparrow is at \((0,0)\), then what is the path the hawk needs to follow? Will the hawk catch the sparrow? The situation is shown in Figure \(\PageIndex{3}\). We pick the path of the sparrow to be along the \(y\)-axis. Therefore, the sparrow is at position \((0, v t)\).

Figure \(\PageIndex{3}\): A hawk at point \((x, y)\) sees a sparrow at point \((0, v t)\) and always follows the straight line between these points.

Solution

First we need the equation of the line of sight between the points \((x, y)\) and \((0, v t)\). Considering that the slope of the line is the same as the slope of the tangent to the path, \(y=y(x)\), we have

\[y^{\prime}=\dfrac{y-v t}{x} \nonumber \]

The hawk is moving at a constant speed, w. Since the speed is related to the time through the distance the hawk travels. We need to find the arclength of the path between \((a, 0)\) and \((x, y)\). This is given by

\[L=\int d s=\int_{x}^{a} \sqrt{1+\left[y^{\prime}(x)\right]^{2}} d x . \nonumber \]

The distance is related to the speed, \(w\), and the time, \(t\), by \(L=w t\). Eliminating the time using \(y^{\prime}=\dfrac{y-v t}{x}\), we have

\[\int_{x}^{a} \sqrt{1+\left[y^{\prime}(x)\right]^{2}} d x=\dfrac{w}{v}\left(y-x y^{\prime}\right) \nonumber \]

Furthermore, we can differentiate this result with respect to \(x\) to get rid of the integral,

\[\sqrt{1+\left[y^{\prime}(x)\right]^{2}}=\dfrac{w}{v} x y^{\prime \prime} \nonumber \]

Even though this is a second order differential equation for \(y(x)\), it is a first order separable equation in the speed function \(z(x)=y^{\prime}(x)\). Namely,

\[\dfrac{w}{v} x z^{\prime}=\sqrt{1+z^{2}} \nonumber \]

Separating variables, we find

\[\dfrac{w}{v} \int \dfrac{d z}{\sqrt{1+z^{2}}}=\int \dfrac{d x}{x} \nonumber \]

The integrals can be computed using standard methods from calculus. We can easily integrate the right hand side,

\[\int \dfrac{d x}{x}=\ln |x|+c_{1} . \nonumber \]

The left hand side takes a little extra work, or looking the value up in Tables or using a CAS package. Recall a trigonometric substitution is in order. [See the Appendix.] We let \(z=\tan \theta \). Then \(d z=\sec ^{2} \theta d \theta\). The methods proceeds as follows:

\[ \begin{aligned} \int \dfrac{d z}{\sqrt{1+z^{2}}}&=\int \dfrac{\sec ^{2} \theta}{\sqrt{1+\tan ^{2} \theta}} d \theta \\[4pt]\ &=\int \sec \theta d \theta \\[4pt] &=\ln (\tan \theta+\sec \theta)+c_{2} \\[4pt] &=\ln \left(z+\sqrt{1+z^{2}}\right)+c_{2} \end{aligned}\label{1.44} \]

Putting these together, we have for \(x>0\),

\[\ln \left(z+\sqrt{1+z^{2}}\right)=\dfrac{v}{w} \ln x+C \nonumber \]

Using the initial condition \(z=y^{\prime}=0\) and \(x=a\) at \(t=0\),

\[0=\dfrac{v}{w} \ln a+C \nonumber \]

or \(C=-\dfrac{v}{w} \ln a\).

Using this value for \(c\), we find

\[ \begin{aligned} \ln \left(z+\sqrt{1+z^{2}}\right) &=\dfrac{v}{w} \ln x-\dfrac{v}{w} \ln a \\[4pt] \ln \left(z+\sqrt{1+z^{2}}\right) &=\dfrac{v}{w} \ln \dfrac{x}{a} \\[4pt] \ln \left(z+\sqrt{1+z^{2}}\right) &=\ln \left(\dfrac{x}{a}\right)^{\dfrac{v}{w}} \\[4pt] z+\sqrt{1+z^{2}} &=\left(\dfrac{x}{a}\right)^{\dfrac{v}{w}} \end{aligned} \nonumber \]

We can solve for \(z=y^{\prime}\), to find

\[y^{\prime}=\dfrac{1}{2}\left[\left(\dfrac{x}{a}\right)^{\dfrac{v}{w}}-\left(\dfrac{x}{a}\right)^{-\dfrac{v}{w}}\right] \nonumber \]

Integrating,

\[y(x)=\dfrac{a}{2}\left[\dfrac{\left(\dfrac{x}{a}\right)^{1+\dfrac{v}{w}}}{1+\dfrac{v}{w}}-\dfrac{\left(\dfrac{x}{a}\right)^{1-\dfrac{v}{w}}}{1-\dfrac{v}{w}}\right]+k \nonumber \]

The integration constant, \(k\), can be found knowing \(y(a)=0 \). This gives

\[ \begin{aligned} 0 &=\dfrac{a}{2}\left[\dfrac{1}{1+\dfrac{v}{w}}-\dfrac{1}{1-\dfrac{v}{w}}\right]+k \\[4pt] k &=\dfrac{a}{2}\left[\dfrac{1}{1-\dfrac{v}{w}}-\dfrac{1}{1+\dfrac{v}{w}}\right] \\[4pt] &=\dfrac{a v w}{w^{2}-v^{2}} \end{aligned}\label{1.46} \]

The full solution for the path is given by

\[y(x)=\dfrac{a}{2}\left[\dfrac{\left(\dfrac{x}{a}\right)^{1+\dfrac{v}{w}}}{1+\dfrac{v}{w}}-\dfrac{\left(\dfrac{x}{a}\right)^{1-\dfrac{v}{w}}}{1-\dfrac{v}{w}}\right]+\dfrac{a v w}{w^{2}-v^{2}} \nonumber \]

Can the hawk catch the sparrow? This would happen if there is a time when \(y(0)=v t\). Inserting \(x=0\) into the solution, we have \(y(0)=\dfrac{a v w}{w^{2}-v^{2}}=v t \). This is possible if \(w>v \).