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A Change of Independent Variable (see Section 2.9) A popular child's toy consists of a small rubber ball attached to a wooden paddle by a rubber band. Assume a 11-oz ball is connected to the paddle by a rubber band having an unstretched length of 2ft2 \mathrm{ft}. When the ball is launched vertically upward by the paddle with an initial speed of 30ft/sec30 \mathrm{ft} / \mathrm{sec}, the rubber band is observed to stretch 1ft1 \mathrm{ft} (to a total length of 3ft3 \mathrm{ft} ) when the ball has reached its highest point. Assume the rubber band behaves like a spring and obeys Hooke's law for this amount of stretching. Our objective is to determine the spring constant kk. (Neglect the weight of the rubber band.)

The motion occurs in two phases. Until the ball has risen 2ft2 \mathrm{ft} above the paddle, it acts like a projectile influenced only by gravity. Once the height of the ball exceeds 2ft2 \mathrm{ft}, the ball acts like a mass on a spring, acted upon by the elastic restoring force of the rubber band and gravity. (a) Assume the ball leaves the paddle at time t=0t=0. Let t2t_2 and t3t_3 represent the times at which the height of the ball is 2ft2 \mathrm{ft} and 3ft3 \mathrm{ft}, respectively, and let mm denote the mass of the rubber ball. Show that an application of Newton's second law of motion leads to the following two-part description of the problem: (i) my=mg,0<t<t2,y(0)=0,y(0)=30m y^{\prime \prime}=-m g, \quad 0<t<t_2, \quad y(0)=0, \quad y^{\prime}(0)=30 (ii) my=k(y2)mgm y^{\prime \prime}=-k(y-2)-m g. t2<t<t3\quad t_2<t<t_3. Here, yy and yy^{\prime} are assumed to be continuous at t=t2t=t_2. We also know that y(t2)=2y\left(t_2\right)=2. y(t3)=3y\left(t_3\right)=3, and y(t3)=0y^{\prime}\left(t_3\right)=0.

If we attempt to solve the problem "directly" with time as the independent variable, it is relatively difficult, since the times t2t_2 and t3t_3 must be determined as part of the problem. Since height y(t)y(t) is an increasing function of time tt over the interval of interest, however, we can view time tt as a function of height, yy, and use yy as the independent variable.

(b) Let v=y=dy/dtv=y^{\prime}=d y / d t. If we adopt yy as the independent variable, the acceleration becomes y=dv/dt=(dv/dy)(dy/dt)=v(dv/dy)y^{\prime \prime}=d v / d t=(d v / d y)(d y / d t)=v(d v / d y). Therefore, (i) mvdvdy=mg,0<y<2,v(0)=30m v \frac{d v}{d y}=-m g, \quad 0<y<2, \quad v(0)=30 (ii) mvdvdy=k(y2)mg,2<y<3m v \frac{d v}{d y}=-k(y-2)-m g, \quad 2<y<3. Here, vv is continuous at y=2y=2 and vy=3=0\left.v\right|_{y=3}=0. Solve these two separable differential equations, impose the accompanying supplementary conditions, and determine the spring constant kk.

Question

The given differential equation has a fundamental set of solutions whose Wronskian W(t)W(t) is such that W(0)=1W(0)=1. What is W(4)W(4) ?

y(4)y+y=0y^{(4)}-y^{\prime \prime}+y=0

Solution

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From Abel’s Theorem\textsf{Abel's Theorem} (see equation (7)(7)) we have that since W(0)=1W(0)=1 we know that the value of the Wronskian of a set of solutions of the differential equation

y(4)y+y=0y^{(4)} - y^{\prime\prime} +y =0

will be:

W(t)=W(0)e0t0ds=1e0=1\pmb{W(t)} = W(0) e^{ -\int_0^t 0 ds } = 1 e^{0 }= \pmb{1 }

We had the constant function 00 in the integral because yy^{\prime\prime\prime} does not appear in the equation !

We conclude that W(4)=1\boxed{ \pmb{ W(4) = 1}}

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