Basic Differential Equations

Basic Differential Equations

Introduction to ODEs and Slope Fields

An Ordinary Differential Equation (ODE) is an equation involving an “unknown” function 𝑦=𝑓(𝑥) and at least one of its derivatives 𝑦′, 𝑦′′, … etc. … . For example, the equation 𝑦 ′ −2𝑥 𝑦 2 =0 A solution to an ODE is a function 𝑦=𝑓(𝑥) that satisfies the differential equation. For example, 𝑦= −1 𝑥 2 is a solution to the differential equation (for 𝑥≠0) above: = 2 𝑥 3 −2𝑥 − 1 𝑥 = 2 𝑥 3 − 2 𝑥 3 𝑦 ′ −2𝑥 𝑦 2 =0 In fact, there are other solutions to this differential equation, such as 𝑦=− 1 𝑥 2 +1 It is often the case that differential equations have infinitely many solutions.

Let’s rewrite the differential equation as 𝑦 ′ =𝑥 𝑦 2
Let’s rewrite the differential equation as 𝑦 ′ =𝑥 𝑦 2 . Suppose we graph the solution 𝑦=− 1 𝑥 2 +1 The differential equation 𝑦 ′ =𝑥 𝑦 2 says that the slope of the graph at the point (𝑥,𝑦) is 𝑥 𝑦 2 .

Suppose we mark the slope at the point (𝑥,𝑦) with a short line segment with slope 𝑥 𝑦 2 .

Example Draw the slope field and visualize some solutions of 𝑦 ′ = 𝑥 2 + 𝑦 2 −1 Solution This is a numerical solution; an exact solution cannot be found.

Let’s look at how to find a general solution to this equation.
𝑦 ′ =𝑥 𝑦 2 We can rewrite the equation in differential form: 𝑑𝑦 𝑑𝑥 =𝑥 𝑦 2 We now separate variables – bring 𝑦 to the left and 𝑥 to the right: 𝑑𝑦 𝑦 2 =𝑥 𝑑𝑥 We can now integrate both sides: 𝑑𝑦 𝑦 2 = 𝑥 𝑑𝑥 − 1 𝑦 = 1 2 𝑥 2 +𝐶 Finally, solve for 𝑦: 𝑦= − 𝑥 2 +𝐶

All the solution curves have the form
𝑦= − 𝑥 2 +𝐶 Except, notice, there is also the solution 𝑦=0 Not all differential equations can be solved this way.

Separable ODEs and Initial Value Problems

The equation 𝑦 ′ =𝑥 𝑦 2 is called a separable differential equation
We say an equation is separable if the variables can be separated: 𝑦 ′ =𝑓 𝑥 𝑔(𝑦) If a differential equation is separable, then it can be solved by integration: 𝑑𝑦 𝑔 𝑦 =𝑓 𝑥 𝑑𝑥 𝑑𝑦 𝑔 𝑦 = 𝑓 𝑥 𝑑𝑥 Once the integration is done, one can (try to) solve for 𝑦 in terms of 𝑥.

Example Visualize and solve the initial value problem
𝑦 ′ =𝑥 𝑦 2 , 𝑦 −3 =1.5 Solution We can visualize an initial value problem by drawing the slope field and the given point (−3,1.5). To solve an initial value problem, we have to find a solution curve that passes through the given point. To solve an initial value problem, we first find a general solution as before: 𝑦= − 𝑥 2 +𝐶 [See slide 8 for solution] Then we use the initial data to figure out what 𝐶 is: 1.5= − −3 2 +𝐶 𝐶=−31/6 So the final answer is: 𝑦= − 𝑥 2 − 31 6 = 6 31−3 𝑥 2

Example (Application to Population Growth)
A bacteria colony with an initial population 5000 undergoes growth at a rate proportional to the population. Suppose after 5 hours the population is found to be 1,000,000. Set up and solve the Initial value problem. Use your solution to predict when the bacteria population will be 1 billion ( 10 9 ). Solution Let 𝑃(𝑡) denote the bacteria population at time 𝑡. Since growth rate is proportional to population 𝑃, we have 𝑑𝑃 𝑑𝑡 =𝑘𝑃 for some (as yet) unknown constant 𝑘 This is a separable equation, which we solve by separating the variables 𝑃 and 𝑡 𝑑𝑃 𝑃 =𝑘 𝑑𝑡 5𝑘= ln 200 𝑘= ln ≈1.0597 ln 𝑃 =𝑘𝑡+𝐶 𝑃= 𝑒 𝑘𝑡+𝐶 Finally, we want to know when 𝑃 𝑡 = 10 9 𝑃=𝐶 𝑒 𝑘𝑡 𝑃 𝑡 =5000 𝑒 𝑡 = 10 9 We use the initial condition 𝑃 0 =5000: 𝑡≈11.52 5000=𝐶 𝑒 0 =𝐶 Then we also have 𝑃 5 =1,000,000 1,000,000=5000 𝑒 𝑘⋅5 200= 𝑒 5𝑘

Example (Application to Constrained Population Growth)
A species of fish is introduced to a pond with an initial population of 𝑃 0 =100 at time 𝑡=0, and satisfies the differential equation 𝑃 ′ =𝑘𝑃 1− 𝑃 𝑀 where, in this case, 𝑘=0.04 and 𝑀=20,000. (a) Solve the initial value problem, and (b) predict when the population will Reach a population of 10,000. (c) Predict the maximum population size. 𝑃 𝑀−𝑃 = 𝑒 𝑘𝑡+𝐶 Solution The equation is separable. = ln 𝑃 𝑀−𝑃 𝑑𝑃 𝑃 1− 𝑃 𝑀 =𝑘 𝑑𝑡 𝑃= 𝑀 𝑒 𝑐+𝑘𝑡 1+ 𝑒 𝑐+𝑘𝑡 So we have Integrate: ln 𝑃 𝑀−𝑃 =𝑘𝑡+𝐶 (∗) 𝑃=𝑀 𝑒 𝑘𝑡 𝑒 −𝑐 + 𝑒 𝑘𝑡 𝑑𝑃 𝑃 1− 𝑃 𝑀 = 𝑘 𝑑𝑡 Since 𝑃 0 = 𝑃 0 =100, we have 𝑃= 𝑒 0.04𝑡 𝑒 0.04𝑡 ln 𝑃 0 𝑀− 𝑃 0 =𝐶 Partial fractions on the left side: (b) Set this equal to 10000 𝑑𝑃 𝑃 1− 𝑃 𝑀 = 𝑃 + 1 𝑀−𝑃 𝑑𝑃 𝐶≈−5.293 𝑒 0.04𝑡 𝑒 0.04𝑡 =10000 Solve (∗) for 𝑃 by exponentiating both sides: = ln 𝑃 − ln (𝑀−𝑃) 𝑡≈132

Example (Application to Constrained Population Growth (continued))
𝑃 𝑡 = 𝑒 0.04𝑡 𝑒 0.04𝑡 (Part c) To find the maximum population, we take the limit as 𝑡→∞: Max Population= lim 𝑡→∞ 𝑒 0.04𝑡 𝑒 0.04𝑡 =20000 This limit could be found using L’Hopital’s rule, or by dividing top and bottom by 𝑒 0.04𝑡 .

Discussion of Constrained Population Growth
We can draw a slope field for the differential equation just discussed 𝑃 ′ =0.04𝑃 1− 𝑃 20000 Recall our solution, 𝑃= 𝑒 0.04𝑡 𝑒 0.04𝑡 We see that solutions behave “like” an exponential function until they inflect and slow down as they approach the maximum population of These solutions are called logistic functions. The population growth model is called the logistic growth model.

Example (Application to Tumor Growth, Numerical Issues)
The growth rate of a tumor decreases exponentially with time, so that the tumor population 𝑃(𝑡) satisfies the differential equation 𝑃 ′ =−𝛼 𝑒 −𝛽𝑡 𝑃, where 𝛼 and 𝛽 are positive constants. (a) Solve the differential equation for a general solution 𝑃(𝑡) (b) On day 0, the tumor population is estimated to be 𝑃 0 =10,000. On day 15, the population is estimated to be Finally, on day 30, the tumor population is estimated to be Find 𝛼, 𝛽, and 𝐶 where 𝐶 is the constant from your general solution to part (a). (c) Calculate the limiting size of the tumor population as 𝑡→∞.

The growth rate of a tumor decreases exponentially with time, so that the tumor population 𝑃(𝑡) satisfies the differential equation 𝑃 ′ =𝛼 𝑒 −𝛽𝑡 𝑃, where 𝛼 and 𝛽 are positive constants. (a) Solve the differential equation for a general solution 𝑃(𝑡) Solution This is a separable equation. 𝑑𝑃 𝑃 =𝛼 𝑒 −𝛽𝑡 𝑑𝑡 Integrate: ln 𝑃 =− 𝛼 𝛽 𝑒 −𝛽𝑡 +𝐶 Solve for 𝑃 𝑃= exp − 𝛼 𝛽 𝑒 −𝛽𝑡 +𝐶

The growth rate of a tumor decreases exponentially with time, so that the tumor population 𝑃(𝑡) satisfies the differential equation 𝑃 ′ =𝛼 𝑒 −𝛽𝑡 𝑃, where 𝛼 and 𝛽 are positive constants. (b) On day 0, the tumor population is estimated to be 𝑃 0 =10,000. On day 15, the population is estimated to be Finally, on day 30, the tumor population is estimated to be Find 𝛼, 𝛽, and 𝐶 where 𝐶 is the constant from your general solution to part (a). Solution We found in part (a) 𝑃= exp − 𝛼 𝛽 𝑒 −𝛽𝑡 +𝐶 If we measure time 𝑡 in days, then the three initial conditions tell us: 10000= exp − 𝛼 𝛽 +𝐶 20000= exp − 𝛼 𝛽 𝑒 −15𝛽 +𝐶 30000= exp − 𝛼 𝛽 𝑒 −30𝛽 +𝐶

10000= exp − 𝛼 𝛽 +𝐶 20000= exp − 𝛼 𝛽 𝑒 −15𝛽 +𝐶 30000= exp − 𝛼 𝛽 𝑒 −30𝛽 +𝐶 The software having numerical difficulties because of the tiny numbers 𝑒 −30𝛽 and large numbers

Suppose we were measuring time 𝑡 in 30-day intervals called “months.”
Then the initial condition at 15 days corresponds to 𝑡=0.5 and 30 days corresponds to 𝑡=1. The equations become: 10000= exp 𝛼 𝛽 +𝐶 20000= exp 𝛼 𝛽 𝑒 −0.5𝛽 +𝐶 30000= exp 𝛼 𝛽 𝑒 −𝛽 +𝐶 These are much more reasonable equations. Indeed, the software has no trouble now! We have found that 𝛼=1.79, 𝛽=1.07, and 𝐶=10.88. 𝑃= exp − 𝛼 𝛽 𝑒 −𝛽𝑡 +𝐶 becomes 𝑃 𝑡 = exp −1.67 𝑒 −1.07𝑡 If we measure 𝑡 in months!

The growth rate of a tumor decreases exponentially with time, so that the tumor population 𝑃(𝑡) satisfies the differential equation 𝑃 ′ =𝛼 𝑒 −𝛽𝑡 𝑃, where 𝛼 and 𝛽 are positive constants. (c) Calculate the limiting size of the tumor population as 𝑡→∞. Solution In part (a) we found that 𝑃= exp − 𝛼 𝛽 𝑒 −𝛽𝑡 +𝐶 And in part (b) we found that that 𝛼=1.79, 𝛽=1.07, and 𝐶=10.88. Taking the limit as 𝑡→∞ of 𝑃(𝑡) and noticing that 𝑒 −𝛽𝑡 →0, we see that lim 𝑡→∞ exp − 𝛼 𝛽 𝑒 −𝛽𝑡 +𝐶 = exp 𝐶 = 𝑒 =53126 We can expect the tumor to approach this size.

Toricelli’s Law Suppose we have a tank full of water with a hole in the bottom. The water drains out according to Toricelli’s Law: velocity= 2𝑔×depth 𝑣= 2𝑔𝑦 Thus the rate of outflow of water is: 𝑑𝑉 𝑑𝑡 =−𝑎𝑣 𝑑𝑉 𝑑𝑡 =−Area×Velocity We also know that 𝑉= 0 𝑦 𝐴 𝑦 𝑑𝑦 Thus 𝑑𝑉 𝑑𝑡 =𝐴 𝑦 𝑑𝑦 𝑑𝑡 Set these two expressions equal to each-other to obtain 𝐴 𝑦 𝑑𝑦 𝑑𝑡 =−𝑎𝑣 =−𝑎 2𝑔𝑦 This is a differential equation we can solve! Toricelli’s Differential Equation 𝐴 𝑦 𝑑𝑦 𝑑𝑡 =−𝑎 2𝑔𝑦

Example A hemispherical tank has top radius 4 ft and at time 𝑡=0 is full of water. At that moment, a circular hole of diameter 1 in. is opened in the bottom. How long will it take for the tank to drain completely? Solution Write Toricelli’s differential equation: 𝐴 𝑦 𝑑𝑦 𝑑𝑡 =−𝑎 2𝑔𝑦 Label the diagram to find 𝐴(𝑦). 𝐴 𝑦 =𝜋 𝑟 2 =𝜋 16− 4−𝑦 2 =𝜋 8𝑦− 𝑦 2 Also 𝑎= 𝜋 and 𝑔=32. The equation becomes 𝜋 8𝑦− 𝑦 2 𝑑𝑦 𝑑𝑡 =−𝜋 𝑦 𝐶= /2 − /2 = 8 𝑦 1/2 − 𝑦 3/2 𝑑𝑦 =− 𝑑𝑡 72 The tank is empty when 𝑦=0, i.e. 16 3 𝑦 3/2 − 2 5 𝑦 5/2 =− 1 72 𝑡+𝐶 0=− 1 72 𝑡 𝑡= ≈2150 𝑠=35min 50s Initial condition: 𝑦 0 =4

16 3 𝑦 3/2 − 2 5 𝑦 5/2 =− 1 72 𝑡 The Clepsydra Problem

Example (Hanging Rope Problem)
A rope of length 𝐿 has ends attached to points ( 𝑥 0 , 𝑦 0 ) and 𝑥 1 , 𝑦 1 and hangs under the Influence of gravity. Write a differential equation that describes the height of the rope 𝑦=𝑦(𝑥). Then solve. Solution Consider a short segment of rope on the interval 𝑥,𝑥+Δ𝑥 . Let 𝜃 be the slope of the rope at 𝑥 and 𝜃+Δ𝜃 be the slope of the rope at 𝑥+Δ𝑥. The left tension force, right tension force, and gravity must add up to zero. We have 𝑇 𝐿 = −𝑇 𝑥 cos 𝜃 ,−𝑇 𝑥 sin 𝜃 𝑇 𝑅 = 𝑇(𝑥+Δ𝑥) cos 𝜃+Δ𝜃 ,𝑇(𝑥+Δ𝑥) sin 𝜃+Δ𝜃 𝐹 𝑔 = 0,−𝑚𝑔 The mass of the rope under consideration is density times the length. The length is approximately Δ𝑥 sec 𝜃 , so 𝑚=𝜌 sec 𝜃 Δ𝑥 Now 𝑇 𝐿 + 𝑇 𝑅 + 𝐹 𝑔 =0, giving us 𝑇 𝑥+Δ𝑥 cos 𝜃+Δ𝜃 −𝑇 𝑥 cos 𝜃 =0 𝑇 𝑥+Δ𝑥 sin 𝜃+Δ𝜃 −𝑇 𝑥 sin 𝜃 −𝜌𝑔 sec 𝜃 Δ𝑥=0 Divide by Δ𝑥 and take the limit as Δ𝑥→0 to find that 𝑑𝑇 𝑑𝑥 cos 𝜃 −𝑇 sin 𝜃 𝑑𝜃 𝑑𝑥 =0 𝑑𝑇 𝑑𝑥 sin 𝜃 +𝑇 cos 𝜃 𝑑𝜃 𝑑𝑥 =𝜌𝑔 sec 𝜃

A rope of length 𝐿 has ends attached to points ( 𝑥 0 , 𝑦 0 ) and 𝑥 1 , 𝑦 1 and hangs under the Influence of gravity. Write a differential equation that describes the height of the rope 𝑦=𝑦(𝑥). Then solve. Solution 𝑑𝑇 𝑑𝑥 cos 𝜃 −𝑇 sin 𝜃 𝑑𝜃 𝑑𝑥 =0 𝑑𝑇 𝑑𝑥 sin 𝜃 +𝑇 cos 𝜃 𝑑𝜃 𝑑𝑥 =𝜌𝑔 sec 𝜃 We analyze the first equation first. Rearranging it, we get 𝑑𝜃 𝑑𝑥 = 1 𝑇 𝑑𝑇 𝑑𝑥 cot 𝜃 tan 𝜃 𝑑𝜃= 𝑑𝑇 𝑇 Integrate both sides: ln sec 𝜃 = ln 𝑇 + 𝐶 1 Exponentiate both sides to find that 𝑇=𝑘 sec 𝜃 Differentiate to find that 𝑑𝑇 𝑑𝑥 =𝑘 sec 𝜃 tan 𝜃 𝑑𝜃 𝑑𝑥 Insert these into the second equation to get: 𝑘 tan 2 𝜃 𝑑𝜃 𝑑𝑥 +𝑘 𝑑𝜃 𝑑𝑥 =𝜌𝑔 sec 𝜃

A rope of length 𝐿 has ends attached to points ( 𝑥 0 , 𝑦 0 ) and 𝑥 1 , 𝑦 1 and hangs under the Influence of gravity. Write a differential equation that describes the height of the rope 𝑦=𝑦(𝑥). Then solve. Solution 𝑘 tan 2 𝜃 𝑑𝜃 𝑑𝑥 +𝑘 𝑑𝜃 𝑑𝑥 =𝜌𝑔 sec 𝜃 This simplifies through tan 2 𝜃 +1= sec 2 𝜃 to sec 𝜃 𝑑𝜃= 𝜌𝑔 𝑘 𝑑𝑥 Integrate both sides: 𝑑𝑦 𝑑𝑥 = tan 𝜃 ln sec 𝜃 + tan 𝜃 = 𝜌𝑔 𝑘 𝑥+ 𝐶 2 This cannot be simplified. Now we figure what 𝑦(𝑥) is, using the fact that 1+ 𝑑𝑦 𝑑𝑥 2 = sec 𝜃 ln 𝑑𝑦 𝑑𝑥 𝑑𝑦 𝑑𝑥 = 𝜌𝑔 𝑘 𝑥+ 𝐶 2 The left-hands ide is just the inverse hyperbolic sine of 𝑑𝑦 𝑑𝑥 . Indeed, sinh −1 𝑢 = ln 𝑢 2 +𝑢 . sinh −1 𝑑𝑦 𝑑𝑥 =𝐴𝑥+𝐵 𝑑𝑦 𝑑𝑥 = sinh (𝐴𝑥+𝐵)

A rope of length 𝐿 has ends attached to points ( 𝑥 0 , 𝑦 0 ) and 𝑥 1 , 𝑦 1 and hangs under the Influence of gravity. Write a differential equation that describes the height of the rope 𝑦=𝑦(𝑥). Then solve. Solution 𝑑𝑦 𝑑𝑥 = sinh (𝐴𝑥+𝐵) It’s also a fact that sinh 𝑢 𝑑𝑢 = cosh 𝑢 +𝐶. Letting 𝑢=𝐴𝑥+𝐵, 𝑑𝑥=𝑑𝑢/𝐴, we have 𝑦= 1 𝐴 cosh 𝐴𝑥+𝐵 +𝐶 The length of this curve from 𝑥= 𝑥 0 to 𝑥= 𝑥 1 is Length=𝐿= 𝑥 0 𝑥 𝑦 ′ 𝑑𝑥 = 𝑥 0 𝑥 sinh 2 (𝐴𝑥+𝐵) 𝑑𝑥 = 𝑥 0 𝑥 1 cosh 𝐴𝑥+𝐵 𝑑𝑥 = 1 𝐴 sinh (𝐴𝑥+𝐵) 𝑥 0 𝑥 1 = 1 𝐴 sinh 𝐴 𝑥 1 +𝐵 − 1 𝐴 sinh 𝐴 𝑥 0 +𝐵

A rope of length 𝐿 has ends attached to points ( 𝑥 0 , 𝑦 0 ) and 𝑥 1 , 𝑦 1 and hangs under the Influence of gravity. Write a differential equation that describes the height of the rope 𝑦=𝑦(𝑥). Then solve. Solution We’ve shown that 𝑦= 1 𝐴 cosh 𝐴𝑥+𝐵 +𝐶 where 𝑦 𝑥 0 = 𝑦 0 𝑦 𝑥 1 = 𝑦 1 1 𝐴 sinh 𝐴 𝑥 1 +𝐵 − 1 𝐴 sinh 𝐴 𝑥 0 +𝐵 =𝐿 This is a system of three equations in the three unknowns 𝐴,𝐵, and 𝐶. For example, if 𝑥 0 , 𝑦 0 =(−5,0), 𝑥 1 , 𝑦 1 =(5,0), 𝐿=15, then the equations become 1 𝐴 cosh −5𝐴+𝐵 +𝐶=0 1 𝐴 cosh 5𝐴+𝐵 +𝐶=0 1 𝐴 sinh 5𝐴+𝐵 − 1 𝐴 sinh −5𝐴+𝐵 =15 The solution is 𝐴=0.324, 𝐵=0, 𝐶=−8.109 𝑦=3.082 cosh 𝑥 −8.109

Example (Falling Satellite Problem)
A satellite is a distance 𝑦=1.390× 10 7 m from the center of the moon, travelling toward the Moon at a speed of v 0 =1200m/s. Predict the time that the satellite will strike the moon, and the speed it will be travelling then. Ignore gravitational effects of all objects except the moon. Solution The radius of the moon is 𝑟 0 =1.737× 10 6 m and we have the following equation from physics: 𝐹=𝑚 𝑎=𝑚 𝑦′′ The only force acting on the satellite is the force of gravity, which is given by 𝐹=− 𝐺𝑀𝑚 𝑦 2 Where 𝑀=7.347× kg is the mass of the moon, 𝑚 the mass of the satellite, 𝐺=6.674× 10 −11 m 3 kg s 2 is the universal gravitational constant, and 𝑦 is the distance from the satellite to the center of the moon. Equating these, we have 𝑚 𝑦 ′′ =− 𝐺𝑀𝑚 𝑦 2 Let 𝑣= 𝑑𝑦 𝑑𝑡 so that 𝑦 ′′ = 𝑑𝑣 𝑑𝑡 = 𝑑𝑣 𝑑𝑦 𝑑𝑦 𝑑𝑡 = 𝑑𝑣 𝑑𝑦 𝑣 𝑣 𝑑𝑣 𝑑𝑦 =− 𝐺𝑀 𝑦 2 This is separable: 𝑣 𝑑𝑣=−𝐺𝑀 𝑑𝑦 𝑦 2

A satellite is a distance 𝑦=1.390× 10 7 m from the center of the moon, travelling toward the Moon at a speed of v 0 =1200m/s. Predict the time that the satellite will strike the moon, and the speed it will be travelling then. Ignore gravitational effects of all objects except the moon. Solution 𝑣 𝑑𝑣=−𝐺𝑀 𝑑𝑦 𝑦 2 Integrate 1 2 𝑣 2 = 𝐺𝑀 𝑦 +𝐶 From the initial condition 𝑣=−1200 when 𝑦=1.390× we find that 𝐶= 1 2 𝑣 0 2 − 𝐺𝑀 𝑦 0 =367,238 Conclude that 𝑣=− 2𝐺𝑀 𝑦 +2𝐶 But 𝑣= 𝑑𝑦 𝑑𝑡 , so this is 𝑑𝑦 𝑑𝑡 =− 2𝐺𝑀 𝑦 +2𝐶

A satellite is a distance 𝑦=1.390× 10 7 m from the center of the moon, travelling toward the Moon at a speed of v 0 =1200m/s. Predict the time that the satellite will strike the moon, and the speed it will be travelling then. Ignore gravitational effects of all objects except the moon. Solution 𝑑𝑦 𝑑𝑡 =− 2𝐺𝑀 𝑦 +2𝐶 This is separable again. − 𝑑𝑦 2𝐺𝑀 𝑦 +2𝐶 =𝑑𝑡 Now we integrate both sides. The integral on the left is difficult, but possible. We do it by computer.

A satellite is a distance 𝑦=1.390× 10 7 m from the center of the moon, travelling toward the Moon at a speed of v 0 = -1200m/s. Predict the time that the satellite will strike the moon, and the speed it will be travelling then. Ignore gravitational effects of all objects except the moon. Solution Thus −2𝑦 𝐶 𝐶+ 𝐺𝑀 𝑦 +𝐺𝑀 ln 𝐺𝑀+2𝐶𝑦+2𝑦 𝐶 𝐶+ 𝐺𝑀 𝑦 𝐶 3/2 =𝑡+ 𝐶 2 The initial conditions tell us that 𝑦=1.390× 10 7 m when 𝑡=0, so we now know that 𝐶 2 =218,871 Impact occurs when 𝑦= 𝑟 0 =1.737× 10 6 m of the moon. 𝑡≈8286s=2 hours, 18 minutes, 6 seconds At this time, we have the velocity of impact 𝑣=− 2𝐺𝑀 𝑦 +2𝐶 ≈−2526m/s

Example (Small (Atmospheric) Rocket Launch Without Air Resistance)
A rocket launches from ground level. The rocket, including the fuel-air mixture inside it, weighs 𝑚 0 . During flight, the fuel burns at a constant rate for 𝜏 seconds. and exhaust gases are ejected from the bottom of the rocket at a rate of 𝛽 kg/s and a speed of 𝑐 (m/s) relative to the rocket. (a) Write a differential equation to describe the height 𝑦 of the rocket. Ignore air resistance, and assume gravity 𝑔 is constant. (b) Solve the differential equation, and find the height 𝑦 and velocity 𝑣 of the rocket after time 𝜏 when the fuel is exhausted.

A rocket launches from ground level. The rocket, including the fuel-air mixture inside it, weighs 𝑚 0 . During flight, the fuel burns at a constant rate for 𝜏 seconds. and exhaust gases are ejected from the bottom of the rocket at a rate of 𝛽 kg/s and a speed of 𝑐 (m/s) relative to the rocket. (a) Write a differential equation to describe the height 𝑦 of the rocket. Ignore air resistance, and assume gravity 𝑔 is constant. Solution The mass of the rocket & fuel at time 𝑡 is 𝑚= 𝑚 0 −𝛽𝑡. Fix a time 𝑡 and consider what happens to the rocket between 𝑡 and 𝑡+Δ𝑡: an amount Δ𝑚 of gas is ejected, as in the diagram: The total momentum at time 𝑡+Δ𝑡 is The total momentum at time 𝑡 is 𝑚𝑣. 𝑚−Δ𝑚 𝑣+Δ𝑣 +Δ𝑚(𝑣−𝑐) =𝑚𝑣−𝑣Δ𝑚+𝑚Δ𝑣−Δ𝑚Δ𝑣+𝑣Δ𝑚−𝑐Δ𝑚 =𝑚𝑣+𝑚Δ𝑣−𝑐Δ𝑚−Δ𝑚Δ𝑣 The change in momentum is therefore 𝑚𝑣+𝑚Δ𝑣−𝑐Δ𝑚 −𝑚𝑣 Δ𝑃=𝑚Δ𝑣−𝑐Δ𝑚−Δ𝑚Δ𝑣 On the other hand, one of Newton’s laws is that 𝐹=Δ𝑃/Δ𝑡. Since the force acting on the rocket is −𝑚𝑔, we have −𝑚𝑔=(𝑚−Δ𝑚) Δ𝑣 Δ𝑡 −𝑐 Δ𝑚 Δ𝑡

A rocket launches from ground level. The rocket, including the fuel-air mixture inside it, weighs 𝑚 0 . During flight, the fuel burns at a constant rate for 𝜏 seconds. and exhaust gases are ejected from the bottom of the rocket at a rate of 𝛽 kg/s and a speed of 𝑐 (m/s) relative to the rocket. (a) Write a differential equation to describe the height 𝑦 of the rocket. Ignore air resistance, and assume gravity 𝑔 is constant. Solution −𝑚𝑔=(𝑚−Δ𝑚) Δ𝑣 Δ𝑡 −𝑐 Δ𝑚 Δ𝑡 Notice that Δ𝑚=𝛽 Δ𝑡, so this becomes −𝑚𝑔= 𝑚−𝛽Δ𝑡 Δ𝑣 Δ𝑡 −𝑐 Δ𝑚 Δ𝑡 Take the limit as Δ𝑡→0 to find that −𝑚𝑔=𝑚 𝑑𝑣 𝑑𝑡 −𝑐𝛽 But the mass remaining at time 𝑡 is 𝑚= 𝑚 0 −𝛽𝑡 so this becomes: − 𝑚 0 −𝛽𝑡 𝑔= 𝑚 0 −𝛽𝑡 𝑑𝑣 𝑑𝑡 −𝑐𝛽 Rearrange: 𝑑𝑣 𝑑𝑡 = 𝑐𝛽 𝑚 0 −𝛽𝑡 −𝑔 This differential equation describes the rocket ∎

A rocket launches from ground level. The rocket, including the fuel-air mixture inside it, weighs 𝑚 0 . During flight, the fuel burns at a constant rate for 𝜏 seconds. and exhaust gases are ejected from the bottom of the rocket at a rate of 𝛽 kg/s and a speed of 𝑐 (m/s) relative to the rocket. (b) Solve the differential equation, and find the height 𝑦 and velocity 𝑣 of the rocket after time 𝜏 when the fuel is exhausted. Solution In part (a) we found the differential equation 𝑑𝑣 𝑑𝑡 = 𝑐𝛽 𝑚 0 −𝛽𝑡 −𝑔 Integrate both sides: 𝑣=−𝑐 ln 𝑚 0 −𝛽𝑡 −𝑔𝑡+𝐶 Clearly 𝑣=0 when 𝑡=0 so we get 𝑐 ln 𝑚 0 =𝐶 Conclude: 𝑣=−𝑐 ln 𝑚 0 −𝛽𝑡 −𝑔𝑡+𝑐 ln 𝑚 0 𝑣=−𝑐 ln 𝑚 0 −𝛽𝑡 𝑚 0 −𝑔𝑡 But 𝑣= 𝑑𝑦 𝑑𝑡 so we get 𝑦= −𝑐 ln 𝑚 0 −𝛽𝑡 𝑚 0 −𝑔𝑡 𝑑𝑡 =− 𝑔 2 𝑡 2 − 𝑐 𝛽 𝑚 0 −𝛽𝑡 + 𝑐 𝛽 𝑚 0 −𝛽𝑡 ln 𝑚 0 −𝛽𝑡 𝑚 𝐶 2

A rocket launches from ground level. The rocket, including the fuel-air mixture inside it, weighs 𝑚 0 . During flight, the fuel burns at a constant rate for 𝜏 seconds. and exhaust gases are ejected from the bottom of the rocket at a rate of 𝛽 kg/s and a speed of 𝑐 (m/s) relative to the rocket. (b) Solve the differential equation, and find the height 𝑦 and velocity 𝑣 of the rocket after time 𝜏 when the fuel is exhausted. Solution Our solution simplifies to 𝑦=𝑐𝑡− 𝑔 2 𝑡 2 + 𝑐 𝑚 0 𝛽 𝑚 0 −𝛽𝑡 𝑚 0 ln 𝑚 0 −𝛽𝑡 𝑚 0 Let 𝑀= 𝑚 0 −𝛽𝑡 𝑚 0 be the fractional mass of the rocket and this becomes: 𝑦=𝑐𝑡− 𝑔 2 𝑡 2 + 𝑐 𝑚 0 𝛽 𝑀 ln 𝑀 𝑣=−𝑐 ln 𝑀 −𝑔𝑡 Let 𝑚 1 = 𝑚 0 −𝛽𝜏 be the final mass of the rocket and we have for final height and velocity: 𝑦 𝜏 =𝑐𝜏− 𝑔 2 𝜏 2 + 𝑐 𝑚 1 𝛽 ln 𝑚 1 𝑚 0 𝑣 𝜏 =−𝑐 ln 𝑚 1 𝑚 0 −𝑔𝜏 ∎

A rocket launches from ground level. The rocket, including the fuel-air mixture inside it, weighs 10kg. During flight, the fuel burns at a constant rate for 𝜏=10 seconds. and exhaust gases are ejected from the bottom of the rocket at a rate of 0.6 kg/s and a speed of 500(m/s) relative to the rocket. Ignoring air resistance and assuming 𝑔=9.8m/ s 2 is constant, find the final height and velocity of the rocket at burnout time 𝜏. Solution From our previous work: 𝑦 𝜏 =𝑐𝜏− 𝑔 2 𝜏 2 + 𝑐 𝑚 1 𝛽 ln 𝑚 1 𝑚 0 𝑣 𝜏 =−𝑐 ln 𝑚 1 𝑚 0 −𝑔𝜏 𝑐=500, 𝑚 1 =10− =4, 𝑚 0 =10, 𝜏=10, 𝛽=0.6, so we get 𝑦 𝜏 = m 𝑣 𝜏 =360.1 m/s

First Order Linear Differential Equations
Every single differential equation we’ve looked at so far has had the form 𝑑𝑦 𝑑𝑥 =𝑓 𝑥 𝑔(𝑦) i.e., have been separable. We now consider a new type of equation, with a completely different solution technique. 𝑑𝑦 𝑑𝑥 +𝑃 𝑥 𝑦=𝑄(𝑥) An equation in this form is called a first order linear equation Notice that 𝑃 and 𝑄 do not have to be linear functions of 𝑥. They can be any (continuous) functions.

Example (Solving a First Order Linear Equation, Edwards & Penney Problem 1.5.16)
Solve the differential equation / Initial value problem 𝑦 ′ = 1−𝑦 cos 𝑥 , 𝑦 𝜋 =2 Solution Rewrite the equation in first order linear form: 𝑦=`1+𝐶 𝑒 − sin 𝑥 𝑑𝑦 𝑑𝑥 + cos 𝑥 𝑦= cos 𝑥 Use the initial condition 𝑦 𝜋 =2 2=1+𝐶 𝑒 −0 =1+𝐶 This is a first order linear differential equation with 𝑃 𝑥 = cos 𝑥 and 𝑄 𝑥 = cos 𝑥 : 𝑑𝑦 𝑑𝑥 +𝑃 𝑥 𝑦=𝑄(𝑥). To solve this equation, we multiply both sides by the integrating factor: 𝐶=1 𝑦=1+ 𝑒 − sin 𝑥 = 𝑒 cos 𝑥 𝑑𝑥 𝐼 𝑥 = 𝑒 𝑃 𝑥 𝑑𝑥 = 𝑒 sin 𝑥 𝑒 sin 𝑥 𝑑𝑦 𝑑𝑥 + cos 𝑥 𝑒 sin 𝑥 𝑦= cos 𝑥 𝑒 sin 𝑥 Notice that the left-hand side can be rewritten: 𝑑 𝑑𝑥 𝑒 sin 𝑥 𝑦 = cos 𝑥 𝑒 sin 𝑥 Integrate both sides with respect to 𝑥 𝑒 sin 𝑥 𝑦= cos 𝑥 𝑒 sin 𝑥 𝑑𝑥 = 𝑒 sin 𝑥 +𝐶 Divide both sides by 𝑒 sin 𝑥

General Solution to First Order Linear Equations
As illustrated in the previous example, the key step needed to solve a first order linear equation is to multiply by the integrating factor 𝐼 𝑥 = 𝑒 𝑃 𝑥 𝑑𝑥 𝑑𝑦 𝑑𝑥 +𝑃 𝑥 𝑦=𝑄(𝑥) 𝐼 𝑥 𝑑𝑦 𝑑𝑥 +𝐼 𝑥 𝑃 𝑥 𝑦=𝐼 𝑥 𝑄(𝑥) Notice that 𝐼 ′ 𝑥 =𝑃 𝑥 𝐼(𝑥). Therefore the left-hand-side can be rewritten (using the product rule): 𝑑 𝑑𝑥 𝐼 𝑥 𝑦 =𝐼 𝑥 𝑄(𝑥) Integrate both sides with respect to 𝑥: 𝐼 𝑥 𝑦= 𝐼 𝑥 𝑄 𝑥 𝑑𝑥 +𝐶 Divide both sides by 𝐼(𝑥) to obtain the final answer! 𝑦= 1 𝐼 𝑥 𝐼 𝑥 𝑄 𝑥 𝑑𝑥 + 𝐶 𝐼 𝑥

Example (Edwards & Penney Problem 1.5.36)
A tank initially contains 60gal of pure water. Brine, containing 1lb of salt per gallon enters the tank at 2 gal/min, and the (perfectly mixed) solution leaves the tank at 3 gal/min; the tank is empty after exactly 1 h. (a) Find the amount of salt in the tank after 𝑡 minutes. (b) What is the maximum amount of salt ever in the tank? Solution The water volume in the tank after 𝑡 minutes is 𝑉=60−𝑡 gal Let 𝐴(𝑡) denote the amount of salt (in lbs) in the tank at time 𝑡 (min) Notice that the concentration 𝑐 of salt in the tank is 𝐴 𝑡 𝑉 = 𝐴 𝑡 60−𝑡 (lbs/gal) Thus, salt is leaving the tank at a rate of 3𝑐= 3𝐴 60−𝑡 (gal/min) On the other hand, salt is entering the tank at a rate of 2 (gal/min) So we have the differential equation 𝑑𝐴 𝑑𝑡 = rate in − rate out =2− 3 60−𝑡 𝐴 Rewrite: 𝑑𝐴 𝑑𝑡 −𝑡 𝐴=2 This is a first-order linear equation with 𝑃 𝑡 = 3 60−𝑡 , 𝑄 𝑡 =2. = −𝑡 3 = 𝑒 −𝑡 𝑑𝑡 Multiply both sides by the integrating factor 𝐼 𝑡 = 𝑒 𝑃 𝑡 𝑑𝑡 = 𝑒 −3 ln 60−𝑡

A tank initially contains 60gal of pure water. Brine, containing 1lb of salt per gallon enters the tank at 2 gal/min, and the (perfectly mixed) solution leaves the tank at 3 gal/min; the tank is empty after exactly 1 h. (a) Find the amount of salt in the tank after 𝑡 minutes. (b) What is the maximum amount of salt ever in the tank? Solution 𝑑𝐴 𝑑𝑡 −𝑡 𝐴=2 𝐼 𝑡 = −𝑡 3 1 60−𝑡 3 𝑑𝐴 𝑑𝑡 −𝑡 4 𝐴= −𝑡 3 𝑑 𝑑𝑡 −𝑡 3 𝐴 = −𝑡 3 1 60−𝑡 3 𝐴= −𝑡 2 +𝐶 𝐴 0 =0 gives 𝐶=− 𝐴= 60−𝑡 − −𝑡 3

A tank initially contains 60gal of pure water. Brine, containing 1lb of salt per gallon enters the tank at 2 gal/min, and the (perfectly mixed) solution leaves the tank at 3 gal/min; the tank is empty after exactly 1 h. (a) Find the amount of salt in the tank after 𝑡 minutes. (b) What is the maximum amount of salt ever in the tank? Solution 𝐴= 60−𝑡 − −𝑡 3 To find the maximum value of 𝐴 for 0≤𝑡≤60 we set 𝑑𝐴 𝑑𝑡 =0. − −𝑡 2 =0 Solve for 𝑡 𝑡≈25.4 or 94.6 The tank is empty after 60s, so the second solution can’t be right. Conclude 𝑡=25.4, and the salt content at that time is: 𝐴 25.4 ≈23.09lbs

Example (Atmospheric Launch with Air Resistance)
We consider the same rocket problem, but now we include air resistance 𝑘=5.4kg/s Write and solve the new differential equation. Find final height and velocity at burnout 𝑡=𝜏. Solution The conservation of linear momentum remains the same, but now 𝐹=−𝑚𝑔−𝑘𝑣. −𝑚𝑔 is the force due to gravity, and the force due to air resistance is −𝑘𝑣 for some constant 𝑘. So instead of −𝑚𝑔=(𝑚−Δ𝑚) Δ𝑣 Δ𝑡 −𝑐 Δ𝑚 Δ𝑡 We now have −𝑚𝑔−𝑘𝑣= 𝑚−Δ𝑚 Δ𝑣 Δ𝑡 −𝑐 Δ𝑚 Δ𝑡 Again take the limit as Δ𝑡→0 and find that −𝑚𝑔−𝑘𝑣=𝑚 𝑑𝑣 𝑑𝑡 −𝑐𝛽 Again, 𝑚= 𝑚 0 −𝛽𝑡 𝑑𝑣 𝑑𝑡 + 𝑘 𝑚 𝑣= 𝑐𝛽 𝑚 −𝑔 𝑑𝑣 𝑑𝑡 + 𝑘 𝑚 0 −𝛽𝑡 𝑣= 𝑐𝛽 𝑚 0 −𝛽𝑡 −𝑔 This is a first order linear differential equation. Multiply both sides by the integrating factor: 𝐼 𝑡 = 𝑒 𝑘 𝑚 0 −𝛽𝑡 𝑑𝑡 = 𝑒 −𝑘 𝛽 ln 𝑚 0 −𝛽𝑡 = 𝑚 0 −𝛽𝑡 −𝑘/𝛽

𝑑 𝑑𝑡 𝑚 0 −𝛽𝑡 −𝑘/𝛽 𝑣 =𝑐𝛽 𝑚 0 −𝛽𝑡 −1−𝑘/𝛽 −𝑔 𝑚 0 −𝛽𝑡 −𝑘/𝛽
Integrate both sides 𝑚 0 −𝛽𝑡 −𝑘/𝛽 𝑣= 𝑐𝛽 𝑘 𝑚 0 −𝛽𝑡 −𝑘/𝛽 + 𝑔 𝛽−𝑘 𝑚 0 −𝛽𝑡 1−𝑘/𝛽 + 𝐶 1 Multiply both sides by 𝑚 0 −𝛽𝑡 𝑘/𝛽 𝑣= 𝑐𝛽 𝑘 + 𝑔 𝛽−𝑘 𝑚 0 −𝛽𝑡 + 𝐶 1 𝑚 0 −𝛽𝑡 𝑘/𝛽 From the initial condition 𝑣=0 when 𝑡=0 we get 𝐶 1 = −𝑐𝛽 𝛽−𝑘 −𝑔 𝑚 0 𝑘 𝑘 𝛽−𝑘 𝑚 0 𝑘/𝛽 Noting that 𝑣=𝑑𝑦/𝑑𝑡 and integrating again, we get: 𝑦= 𝑐𝛽 𝑘 𝑡− 𝑔 2𝛽 𝛽−𝑘 𝑚 0 −𝛽𝑡 2 − 𝐶 1 𝛽 1+𝑘/𝛽 𝑚 0 −𝛽𝑡 1+𝑘/𝛽 + 𝐶 2 From the initial condition 𝑦=0 when 𝑡=0 we get 𝐶 2 = 𝑔 𝑚 𝛽 𝛽−𝑘 + 𝐶 1 𝑚 0 1+𝑘/𝛽 𝛽 1+𝑘/𝛽

𝑦= 𝑐𝛽 𝑘 𝑡− 𝑔 2𝛽 𝛽−𝑘 𝑚 0 −𝛽𝑡 2 − 𝐶 1 𝛽 1+𝑘/𝛽 𝑚 0 −𝛽𝑡 1+𝑘/𝛽 + 𝐶 2
𝑣= 𝑐𝛽 𝑘 + 𝑔 𝛽−𝑘 𝑚 0 −𝛽𝑡 + 𝐶 1 𝑚 0 −𝛽𝑡 𝑘/𝛽 𝐶 1 = −𝑐𝛽 𝛽−𝑘 −𝑔 𝑚 0 𝑘 𝑘 𝛽−𝑘 𝑚 0 𝑘/𝛽 𝐶 2 = 𝑔 𝑚 𝛽 𝛽−𝑘 + 𝐶 1 𝑚 0 1+𝑘/𝛽 𝛽 1+𝑘/𝛽 The final velocity 47.4m/s and height 354m at burnout are significantly smaller than before. Clearly, air resistance cannot be ignored.

Example (Edwards & Penney, Example 1.7.6)
Suppose a flexible 4-ft rope starts with 3 ft of its length arranged in a heap at the edge of a high horizontal table, with the remaining 1 foot hanging (at rest) off the table. At time 𝑡=0 the heap begins to unwind and the rope begins to gradually fall off the table, under the force of gravity pulling on the overhanging part. Under the assumption that frictional forces of all sorts are negligible, how long will it take for all the rope to fall off the table?

Example (Edwards & Penney, Example 1.7.6)
Suppose a flexible 4-ft rope starts with 3 ft of its length arranged in a heap at the Edge of a high horizontal table, with the remaining 1 foot hanging (at rest) off the table. At time 𝑡=0 the heap begins to unwind and the rope begins to gradually fall off the table, under the force of gravity pulling on the overhanging part. Under the assumption that frictional forces of all sorts are negligible, how long will it take for all the rope to fall off the table? Solution Let 𝑥(𝑡) denote the length of overhanging rope at time 𝑡 and 𝑣(𝑡) denote the velocity then with positive direction pointing downward. So 𝑥 0 =1, 𝑣 0 =0, and 𝑣 𝑡 =𝑥′(𝑡). Let 𝜌 be the linear density of the rope, so that the overhanging rope has mass 𝑚=𝜌𝑥. The force of gravity 𝐹 is 𝑚𝑔=𝜌𝑔𝑥. Newton’s second law then says 𝐹=𝜌𝑔𝑥= 𝑑 𝑑𝑡 𝑚𝑣 𝜌𝑔𝑥=𝑣 𝑑𝑚 𝑑𝑡 +𝑚 𝑑𝑣 𝑑𝑡 𝜌𝑔𝑥=𝜌𝑣 𝑑𝑥 𝑑𝑡 +𝜌𝑥 𝑑𝑣 𝑑𝑡 𝑔𝑥= 𝑣 2 +𝑥 𝑑𝑣 𝑑𝑡 The rope falls off the table in 0.54s. This is a system of equations: 𝑔𝑥= 𝑣 2 +𝑥 𝑑𝑣 𝑑𝑡 𝑥 0 =1 𝑣= 𝑑𝑥 𝑑𝑡 𝑣 0 =0

We will now solve the previous problem exactly, without using NDSolve or DSolve.
Warning: the following solution is extremely clever. It is recommended that the reader don a heat-resistance face-mask to avoid the face-melting awesomeness of the following solution. 𝑔𝑥= 𝑣 2 +𝑥 𝑑𝑣 𝑑𝑡 Since 𝑣=0 when 𝑥=1, we get 𝐶=−𝑔/3≈−3.27 Rearrange: Write 𝑣=𝑑𝑥/𝑑𝑡. 1 2 𝑥 𝑑𝑥 𝑑𝑡 2 − 𝑔 3 𝑥 3 =− 𝑔 3 𝑥 𝑑𝑣 𝑑𝑡 + 𝑣 2 −𝑔𝑥 =0 Notice by the chain rule that 𝑑𝑣 𝑑𝑡 = 𝑑𝑣 𝑑𝑥 𝑑𝑥 𝑑𝑡 = 𝑑𝑣 𝑑𝑥 𝑣 This is a separable equation: solve for 𝑑𝑥/𝑑𝑡 𝑑𝑥 𝑑𝑡 = 2𝑔 𝑥 3 −1 𝑥 2 𝑥𝑣 𝑑𝑣 𝑑𝑥 + 𝑣 2 −𝑔𝑥 =0 Multiply both sides by 𝑥. 𝑥 𝑑𝑥 𝑥 3 −1 = 2𝑔 3 𝑑𝑡 Let 𝑇 be the answer, so 𝑥 2 𝑣 𝑑𝑣 𝑑𝑥 + 𝑥 𝑣 2 −𝑔 𝑥 2 =0 𝑥 𝑇 =4 𝑥=1 𝑥=4 𝑥 𝑑𝑥 𝑥 3 −1 = 𝑡=0 𝑡=𝑇 2𝑔 3 𝑑𝑡 Notice that the left-hand side can be rewritten: 𝑑 𝑑𝑥 𝑥 2 𝑣 2 − 𝑔 3 𝑥 3 =0 𝑇= 3 2𝑔 𝑥 𝑑𝑥 𝑥 3 −1 =0.541s Integrate both sides: 1 2 𝑥 2 𝑣 2 − 𝑔 3 𝑥 3 =𝐶 (The last integral was computed numerically. Note 𝑔=32ft/ s 2 )

Alternative Solution to the Falling Rope Problem
In the last two solutions to the falling rope problem, we used Newton’s law: 𝐹= 𝑑 𝑑𝑡 𝑚𝑣 What if we use Conservation of Energy instead? Rearrange: 𝑃 𝐸 0 +𝐾 𝐸 0 =𝑃 𝐸 𝑡 +𝐾 𝐸 𝑡 𝑥 𝑣 2 +𝑔 1− 𝑥 2 =0 Solve for 𝑣 and note 𝑣=𝑑𝑥/𝑑𝑡. Suppose we say the table is at a height 0, so the hanging rope has negative potential energy because it’s at a negative height. Let’s compute the potential energy of the rope when a length 𝑥 is hanging. 𝑑𝑥 𝑑𝑡 = 𝑔( 𝑥 2 −1) 𝑥 𝑃𝐸=𝑃 𝐸 on table +𝑃 𝐸 hanging 𝑃𝐸=𝜌𝑔 4−𝑥 ⋅0+𝜌𝑔𝑥 − 𝑥 2 =− 𝜌𝑔 2 𝑥 2 This is separable. 𝑥 𝑥 2 −1 𝑑𝑥= 𝑔 𝑑𝑡 For kinetic energy, the rope on the table is at rest and has none. The hanging rope is moving at velocity 𝑣: 𝑥 𝑥 2 −1 𝑑𝑥 = 0 𝑇 𝑔 𝑑𝑡 𝐾𝐸=𝐾 𝐸 on table +𝐾 𝐸 hanging 𝐾𝐸= 𝜌𝑥 𝑣 2 = 𝜌 2 𝑥 𝑣 2 𝑇= 1 𝑔 𝑥 𝑥 2 −1 𝑑𝑥 By conservation of energy, 𝑃𝐸+𝐾𝐸 is constant. The initial energy is − 𝜌𝑔 so we obtain 𝑇=0.488 𝜌 2 𝑥 𝑣 2 − 𝜌𝑔 2 𝑥 2 =− 𝜌𝑔 2 This answer is different than the one we got before. Why?

Basic Differential Equations

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Basic Differential Equations

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