Differential Equations Tutor, Volume I Worksheet 1 What is ... · derivative. Then, this is a...

Differential Equations Tutor, Volume I

Worksheet 1

What is a Differential Equation?

Worksheet for Differential Equations Tutor, Volume I, Section 1:

What is a Differential Equation?

For each of the following equations, determine: Is it or is it not a differential equation?

If it is a differential equation, is it an ordinary differential equation or a partial differential

equation? If it is a differential equation, what is its order? Explain all your reasoning.

1. The Malthusian model for population growth: dPdt

= bP where P is the population at a

given time and b is a constant representing the growth rate.

2. Einstein’s equation for energy: E = mc2.

3. The Black-Scholes Equation, which is important in finance: ∂V∂t

+ 12σ2S2 ∂2V

∂S2 + rS ∂V∂S−

rV = 0.

c©2019 MathTutorDVD.com 1

4. The equation ddx

cos 2x = −2 sin 2x.

5. The logistic equation, which has applications in all sorts of scientific fields: dydx

=

y(1− y).

6. The heat equation, describing the flow of heat: ∂u∂t

= α(∂2u∂x2

+ ∂2u∂y2

+ ∂2u∂z2

).


7. The equation∫∫

xdxdt =∫ydt. Hint: What does this equation become under differ-

entiation?

8. The equation ddyey = ey.

9. The equation describing the motion of a spring under damped conditions, md2xdt2

+

γ dxdt

+ kx = 0.


10. The definition of a Laplacian function, ∂2ψ∂x2

+ ∂2ψ∂y2

= 0.

11. Faraday’s Law, which states that ∇ × E = −∂B∂t

where E,B are the electric and

magnetic fields respectively.

12. The equation d3ydt3

=(dydt

)2+ y.


13. The Euler equation for incompressible fluid flow, ∂u∂x

+ ∂v∂y

= 0.

For each of the following solutions to differential equations, determine: is it a general

solution or a specific solution? If it is a general solution, find a specific solution compatible

with the given initial conditions. If it is a specific solution, determine whether or not it is

compatible with the given initial conditions.

14. x(t) = 5t−1 for the initial condition x(0) = −1. This differential equation solves dxdt

= 5.

15. x(t) = t+C for the initial condition x(1) = 3. This differential equations solves dxdt

= 1.


16. x(t) = Cet for x(0) = 2. This differential equation solves dxdt

= x.

17. x(t) = 3e2t for x(1) = 3e2. This differential equation solves dxdt

= 2x.

18. x(t) = 5e−t for x(0) = 3. This differential equation solves dxdt

= −x.


19. x(t) = a2t2+100 for the initial conditions x(0) = 100, x′(0) = 0. This differential equation

solves d2xdt2

= a for a constant.

20. x(t) = a2t2 + C1t + C2 for the initial conditions x(0) = 50, x′(0) = 0. This differential

equation also solves d2xdt2

= a for a constant.

21. x(t) = 9t2 + 50t + 5 for the initial conditions x(0) = 5, x′(0) = 10. This differential

equation solves d2xdt2

= 18.


22. x(t) = 6t2 + C1t + C2 for the initial conditions x(1) = 5, x′(1) = 10. This differential


= 12.

23. x(t) = 4t2 +C1t+C2 for the initial condition x(1) = 2. This differential equation solves

d2xdt2

= 8. Hint: there may be more than one specific solution compatible with the initial

condition.

24. x(t) = a2t2 +5t+C for the initial condition x(2) = −3. This differential equation solves

dxdt

= at+ 5 for a constant.


What do differential equations actually mean? Answer the following questions about the

radioactive decay equation, dQdt

= −rQ(t) where Q is the quantity of radioactive material

and r is the positive rate of decay.

25. Suppose that Q(t) is positive. Is the amount of radioactive material then increasing

or decreasing?

26. As the amount of radioactive material decreases, does the rate of decrease go up or

down (in magnitude)? Explain your reasoning.

27. The change in the rate of change is given by d2Qdt2

. Is this quantity positive or negative?

Hint: take the derivative of dQdt

= −rQ(t).


Now we will examine more closely the equation F = ma or d2xdt2

= Fm

.

28. The force of gravity on an object of mass m is −mg, where g is Earth’s gravitational

constant in meters per second squared and the negative sign is because the object

is “falling.” What is the acceleration due to gravity on an object?

29. What is the velocity at time t of an object moving only under the influence of gravity, if

its initial velocity is zero? Is the velocity dependent on the initial height? Why or why

not?


30. What is the velocity at time t of an object moving only under the influence of gravity,

if its initial velocity is v0? Is the velocity dependent on the initial height? Why or why

not? Do the units check out in this equation for velocity?

31. What is the position at time t of an object moving only under the influence of gravity,

if the initial height is h and the initial velocity is zero?


if the initial height is h and the initial velocity is v0? Is the position dependent on the

initial velocity? Why or why not? Do the units check out in this equation for position?

33. An object falls out of an helicopter that is flying at 1000m altitude. How long does


it take the object to reach the ground? Assume that g = 9.8ms2

. Also assume no air

resistance or other effects.

34. An object is thrown out of a helicopter that is flying at 500m altitude. The object

is thrown upwards at 10ms

. How long does it take the object to reach the ground?

Assume that g = 9.8ms2

. Also assume no air resistance or other effects.


Answer key.

Is it a differential equation?

1. The Malthusian model for population growth: dPdt

= bP where P is the population at a

given time and b is a constant representing the growth rate.

This is a simple differential equation that is basic and well-known in science. The

change of a population P with time t is directly related to the population P . In

other words, the differential dPdt

is a function of P - a linear function, in fact. If

P is greater, its change is greater; and if P is lesser, its change is lesser. The

growth in population is proportional to the population that already exists - for in-

stance, the more pairs of breeding rabbits there are on an island, the more the

rabbits will reproduce as long as resources such as space remain unlimited. It is

also a ordinary differential equation, since there is only one variable and it is chang-

ing with respect to time. All of the derivatives are “clean” and not partial. Finally,

its order is 1, since the derivative taken is a first derivative. Therefore, this is an

Answer: ordinary differential equation of order 1 . This simple ordinary differen-

tial equation of degree 1 is well within the scope of this course; we will learn that in

fact the solution to this equation is

P (t) = P0ebt

where P0 is the initial population.


2. Einstein’s equation for energy: E = mc2.

There is no derivative or differential in this very important equation. The speed of

light is constant and the mass of a system is constant, so the energy also remains

constant. It is an equivalence between mass and energy, or a formula to calculate one

quantity given the other. Therefore this equation is Answer: not a differential equation

and it is outside of the scope of differential equations.


3. The Black-Scholes Equation, which is important in finance: ∂V∂t

+ 12σ2S2 ∂2V

∂S2 + rS ∂V∂S−

rV = 0.

This is a differential equation; in fact, the changes in time of V (the price of an op-

tion) and S (the price of a stock) are given. It is a partial differential equation, since

there are two variables in play here. The order is 2, since there is a second partial

derivative. Then, this is a Answer: partial differential equation of order 2 . Partial

differential equations are more difficult to solve than ordinary differential equations

and are more advanced than the topics of this course. In fact, for the Black-Scholes

equation, there is no general numerical solution. Rather, given initial conditions, com-

puter numerical approximations can be used to model the behavior of the functions

V and S, even if no closed-form solution for these functions can be found.


4. The equation ddx

cos 2x = −2 sin 2x.

This is an identity stating that the derivative (instantaneous slope) of the function

cos 2x is −2 sin 2x. It is already, as it were, a “solved problem;” there is no differential

equation here to solve. The term x is a variable, not a function, so it is not the subject

of a differential equation. It is merely allowed to vary freely, and no constraints are

placed upon its varying or upon its derivative. Therefore, although there is a derivative

in this equation, it is Answer: not a differential equation since the differential is

not tied to the behavior of a variable.


5. The logistic equation, which has applications in all sorts of scientific fields: dydx

=

y(1− y).

Here, y is a function of x - indicated by the fact that y is differentiated with respect to

x. Not every function y(x) will satisfy this equation, however. Consider y(x) = 12, a

constant function. Then,dy

dx= 0

but

y(1− y) = 1

2

(1− 1

2

)=

1

4

so the differential equation is not satisfied. The function y(x) = x also does not satisfy

this equation: we havedy

dx= 1

so

1 = x(1− x)

which is not true for any real x. Even if it were true for some real x, it would not be

a solution to the differential equation as it would not hold for all x. The point of this

differential equation is to find a function y(x) so that

dy

dx= y(x) (1− y(x))

at all x, not just for some particular x. It is an equation tying the behavior of the func-

tion y(x) to its derivative, hence a basic differential equation to which the solution will

be some function y(x). It is a ordinary differential equation since all the functions and

derivatives are in terms of one variable, and its order is 1 since the highest derivative

taken is a first-order derivative.

Therefore, Answer: this is an ordinary differential equation of order 1 . The so-


lution turns out to be of moderate difficulty; in fact, the general solution is

y(x) =1

1 + Ce−x

for any constant C. It is used in many fields of science; in fact, it can be used to

analyze with greater accuracy and refinement the same growing population that the

Malthusian differential equation can study. The greater complexity of the equation

and solution is due to the greater accuracy of the underlying model.


6. The heat equation, describing the flow of heat: ∂u∂t

= α(∂2u∂x2

+ ∂2u∂y2

+ ∂2u∂z2

).

Here, the change of the heat variable u(x, y, z, t) with respect to time in a given posi-

tion (x, y, z) is modeled by the local flow of temperature at x, y, z. The goal is to de-

termine what the temperature is as a function of position and time, and this equation

is the condition placed on the evolution of the temperature function. The temperature

function, whatever it may be, is known to change in this way. What then is the tem-

perature function, given these constraints on how it must change in time and space?

This is a more advanced differential equations problem, because it is a partial differ-

ential equation (many different variables) and its order is 2 because there are second

derivatives. It is a Answer: partial differential equation of order 2 . As such, it is

beyond the scope of this introductory course in differential equations. In higher-level

study of differential equations, it turns out that there is often no closed-form solution;

that is, there is no function u(x, y, z, t) that can be cleanly expressed, but this differ-

ential equation allows a computer to model the function u as it evolves, so values at

a specific time can be determined.


7. The equation∫∫

xdxdt =∫ydt. Hint: What does this equation become under differ-

entiation?

Strictly speaking, this is an integral equation; however, it can be easily transformed

into a differential equation. If ∫∫xdxdt =

∫ydt

thend

dt

∫∫xdxdt =

d

dt

∫ydt

By the fundamental theorem of calculus, then,∫xdx = y

Once we differentiate again with respect to x, we have that

dy

dx= x

The variable y is a function of x, and the relationship is given by a derivative, so this

is a differential equation. It is ordinary and its order is 1, since the derivative is only

taken once. Therefore it is an Answer: ordinary differential equation of order 1 .

The solution is

y(x) =x2

2+ C

as follows from Calculus I, where both variables also may be functions in terms of t.


8. The equation ddyey = ey.

Although a derivative is taken here, this is an identity - it is something that is always

true. In Calculus I, we learned that

d

dyey = ey

for any variable y. This is true because y is a variable. It is completely free and

there are no constraints placed upon it or its differential by this identity. Then, this is

Answer: not a differential equation.


9. The equation describing the motion of a spring under damped conditions, md2xdt2

+

γ dxdt

+ kx = 0.

Here, x is the position of the spring. The position of the spring is related to its ac-

celeration and its velocity - that is, to d2xdt2

and dxdt

respectively. In other words, this is

a differential equation, and it is ordinary because there is only one variable and no

partial derivatives. The order is 2 because of the presence of the second derivative.

Then, this is an Answer: ordinary differential equation of order 2 . The fact that

the order is 2 makes the problem more complicated and its study will have to wait

for the second half of this course. The solution to this differential equation will give

the motion of a spring with respect to time. In fact, the solution depends on whether

γ2 − 4mk is less than zero, greater than zero, or equal to zero. If γ2 − 4mk < 0 there

is a trigonometric solution. If γ2 − 4mk > 0 the solution is exponential. Finally, if

γ2 − 4mk = 0, there is a more complicated exponential soultion.


10. The definition of a Laplacian function, ∂2ψ∂x2

+ ∂2ψ∂y2

= 0.

Some function ψ(x, y) is given a constraint in terms of its derivatives. A function ψ

that satisfies this equation is known as a Laplacian function. Laplacian functions are

important in physics and computer science. This is a differential equation, but since it

is partial (that is, multiple variables are at play) and of order 2, it is beyond the scope

of this course. Answer: a partial differential equation of order 2 .


11. Faraday’s Law, which states that ∇× E = −∂B∂t

where E,B are the electric and mag-

netic fields respectively.

Faraday’s Law describes the relationship between the electric and magnetic fields.

Both fields are defined in three dimensions, so the curl ∇× E is well-defined:

∇× E =

(∂

∂yEz −

∂

∂zEy

)ı−

(∂

∂xEz −

∂

∂zEx

)+

(∂

∂xEy −

∂

∂yEx

)where Ex, Ey, Ez are the components of E in the directions x, y, z respectively. Since

∂B

∂t=

∂

∂tBxı+

∂

∂tBy+

∂

∂tBzk

where likewise Bx, By, Bz are the components of B, this equation is in fact equal to

three differential equations:

∂

∂yEz −

∂

∂zEy = −

∂

∂tBx

∂

∂xEz −

∂

∂zEx =

∂

∂tBy

∂

∂xEy −

∂

∂yEx = −

∂

∂tBz

This system of differential equations characterizes the change in the electric field with

respect to time, space, and the change of the magnetic field. It is a differential equa-

tion, but the presence of partial derivatives makes it a partial differential equation that

is outside the scope of this course. Its order is 1 so it is a

Answer: partial differential equation of order 1 . If the magnetic field evolves, cor-

responding changes in the electric field - a process known as electromagnetic induc-

tion, which can power generators - can be derived through this equation.


12. The equation d3ydt3

=(dydt

)2+ y.

The function y is related to its change in time, which is a basic ordinary differential

equation. Its order is 3, since a third derivative is taken. This is an

Answer: ordinary differential equation of order 3 . Even though it is ordinary, it

eludes easy solution due to the high order and the presence of the square in the

exponent. Numerical solutions can be found based on the differential equation, but it

is hard to find a closed-form solution.


13. The Euler equation for incompressible fluid flow, ∂u∂x

+ ∂v∂y

= 0.

The order is 1 since only one derivative is taken, but the derivatives that are taken are

partial. Then, this equation - which models the flow of fluids or of air - is outside of the

scope of this course. This is a Answer: partial differential equation of order 1 .


In the following problems, we determine if the following correct solutions to differential

equations are general or specific, and we either find the specific solutions or check the

ones provided for accuracy.

14. x(t) = 5t−1 for the initial condition x(0) = −1. This differential equation solves dxdt

= 5.

The differential equation dxdt

= 5 has a whole family of solutions of the form

x(t) = 5t+ C

for some constant C. Then, 5t+ C is the general solution. One specific solution is

x(t) = 5t− 1

We want to know if this specific solution is compatible with the condition x(0) = −1.

We have

x(0) = 5 · 0− 1 = −1

so it is compatible with the initial condition. Then, this is the

Answer: specific solution that is consistent with the given initial conditions . The

specific solution x(t) = 5t+2 would not satisfy this initial condition but only some other

initial condition.


15. x(t) = t+C for the initial condition x(1) = 3. This differential equations solves dxdt

= 1.

The differential equation dxdt

= 1 is solved by any equation of the form

x(t) = t+ C

Since C need not be specified to know that this is a solution, this is a general solu-

tion. We are given the initial condition that

x(1) = 3

which means that

3 = 1 + C

so

C = 2

The correct specific solution, then, is Answer: x(t) = t+ 2 .


16. x(t) = Cet for x(0) = 2. This differential equation solves dxdt

= x.

Any solution of the form x(t) = Cet satisfies the differential equation

dx

dt= x(t)

so this is a general solution, since the constants are not specified. The specific

solution will have

x(0) = 2

We know that

x(0) = Ce0 = C

so we have

C = 2, Answer: x(t) = 2et

as the only solution - the specific solution - that meets this initial condition.


17. x(t) = 3e2t for x(1) = 3e2. This differential equation solves dxdt

= 2x.

The differential equationdx

dt= 2x(t)

is one that we will learn to solve in this course. The general solution is

x(t) = Ce2t

for any constant C. We have been provided with the specific solution

x(t) = 3e2t

which does indeed satisfy

x(1) = 3e2(1) = 3e2

Then, it is the Answer: specific solution consistent with the given initial conditions .


18. x(t) = 5e−t for x(0) = 3. This differential equation solves dxdt

= −x.

The general solution

x(t) = Ce−t

solves the differential equationdx

dt= −x

and the specific solution satisfying the initial condition

x(0) = 3

has

C = 3, x(t) = 3e−t

The specific solution that we have been given, x(t) = 5e−t, does NOT satisfy x(0) = 3,

so it is Answer: inconsistent with the initial condition.


19. x(t) = a2t2 + 100 for the initial conditions x(0) = 100, x′(0) = 0. This differential equa-

tion solves d2xdt2

= a for a constant.

The differential equation d2xdt2

= a describes, say, the behavior of an object near a

planet with gravitational constant −a. The general solution is

x(t) =a

2t2 + C0t+ C1

Here we have been given the specific solution

C0 = 0, C1 = 100

so the specific solution is

x(t) =a

2t2 + 100

It does indeed satisfy

x(0) = 100

How about x′(0)? We can take the derivative of the specific solution to get

x′(t) = at

as the specific solution for velocity (the general solution would be x′(t) = at + C0).

This does indeed satisfy

x′(0) = 0

so it Answer: is a specific solution that is consistent with the given initial conditions

- representing an object dropped from 100 units of height in a gravitational field with

gravitational constant −a (for a negative).


20. x(t) = a2t2 + C1t + C2 for the initial conditions x(0) = 50, x′(0) = 0. This differential

equation also solves d2xdt2

= a for a constant.

In general, the differential equation d2xdt2

= a has solution

x(t) =a

2t2 + C1t+ C2

We have been given constraints; that the initial position is 50 and the initial velocity is

zero. We have

x(0) =a

2(0) + C1(0) + C2

so the initial condition is satisfied when

C2 = 50

Given the equation for position, the velocity is

x′(t) = at+ C1

so the condition that the initial velocity is zero implies that

x′(0) = a(0) + C1 = 0

so

C1 = 0

and the specific solution is

Answer: x(t) =a

2t2 + 50


21. x(t) = 9t2 + 50t + 5 for the initial conditions x(0) = 5, x′(0) = 10. This differential


= 18.

Here, we consider the particular value a = 18. The general solution to d2xdt2

is then

x(t) = 9t2 + C1t+ C2

for constants C1, C2 that fit with the initial conditions. The specific solution we have

been given is

x(t) = 9t2 + 50t+ 5

This specific solution has

x(0) = 9(0)2 + 50(0) + 5 = 5

as desired. The specific solution has

x′(t) = 18t+ 50

which, however, does NOT correspond to the initial condition x′(0) = 10 because

for this specific solution we have x′(0) = 50. The specific solution that we have been

given is then Answer: inconsistent with the initial condition. So the correct specific

solution for these initial conditions is

x(t) = 9t2 + 10t+ 5

and the specific solution that we have been given corresponds to the initial conditions

x(0) = 5, x′(0) = 50.


22. x(t) = 6t2 + C1t + C2 for the initial conditions x(1) = 5, x′(1) = 10. This differential


= 12.

Every solution to d2xdt2

= 12 is of the form

x(t) = 6t2 + C1t+ C2

We seek to find C1, C2 so that the initial conditions are satisfied. We have that

x(1) = 6 + C1 + C2 = 5

so

C1 + C2 = −1

The velocity equation is

x′(t) = 12t+ C1

so the initial velocity condition tells us that

x′(1) = 12 + C1 = 10

so

C1 = −2

which implies

C2 = 1

Then, the specific solution copmatible with these initial conditions is

Answer: x(t) = 6t2 − 2t+ 1


23. x(t) = 4t2 +C1t+C2 for the initial condition x(1) = 2. This differential equation solves

d2xdt2

= 8. Hint: there may be more than one specific solution compatible with the initial

condition.

Every solution to d2xdt2

= 8 is of the form

x(t) = 4t2 + C1t+ C2

so this is an general solution. We have that

x(1) = 4 + C1 + C2 = 2

as our initial condition, so

C1 + C2 = −2

Any specific solution that satisfies this equation will match our initial condition. One

specific solution, then, is

C1 = 0, C2 = −2

which gives the specific solution Answer: x(t) = 4t2 − 2 . Other specific solution that

match the initial condition are equally correct - for instance,

x(t) = 4t2 + t− 3

or in general

x(t) = 4t2 + Ct− (C + 2)

for any value C.


24. x(t) = a2t2 +5t+C for the initial condition x(2) = −3. This differential equation solves

dxdt

= at+ 5 for a constant.

The general solution to the differential equation

dx

dt= at+ 5

is

x(t) =a

2t2 + 5t+ C

but this can be made specific to match the initial conditions. If x(2) = −3, we have

−3 =a

2(2)2 + 5(2) + C

so

C = −3− 2a− 10 = −13− 2a

and the specific solution is Answer: x(t) =a

2t2 + 5t− 13− 2a .


Explaining the meaning of differential equations:

25. In dQdt

= −rQ(t), suppose that Q(t) is positive. Is the amount of radioactive material

then increasing or decreasing?

Suppose that Q(t) is positive - that is, that there is a positive amount of radioactively

decaying material. Then, dQdt

= −rQ(t) is negative since r and Q(t) are both positive.

Answer: Q(t) > 0 =⇒ dQ

dt< 0 .

This means that the amount of radioactive material is decreasing, since its derivative

in time is negative. This fits with the physical understanding of the process of radioac-

tive decay. As some radioactive material decays, the amount of radioactive material

goes down. The amount of decrease at any time is related to the amount present at

that time and the constant rate of decay. This relation between the rate of change

and the quantity present sets up a differential equation.


26. As the amount of radioactive material decreases, does the rate of decrease go up or

down (in magnitude)? Explain your reasoning.

Suppose that at the beginning, Q(t) = 10 and at some point later Q(t) = 5. The rate

of change when Q(t) = 10 is −10r, but when Q(t) = 5 the rate of change is only −5r.

The decrease, then, continues to slow and slow.

Answer: As Q(t) decreases,dQ

dtalso decreases in magnitude.

If the rate of decrease were a constant dQdt

= −C, then the function Q(t) = −Ct + C0

would characterize the quantity of radioactive substance, and the quantity would hit

zero at t = C0

C. However, the rate of decrease slows as the amount of substance de-

creases, so it takes much more time for the substance to be exhausted. In fact, rather

than a linear function, Q(t) decays exponentially. The only function that satisfies the

differential equation indicating the rate of decay is an exponential function, as we will

explore in subsequent lessons.


27. The change in the rate of change is given by d2Qdt2

. Is this quantity positive or negative?

Hint: take the derivative of dQdt

= −rQ(t).

The rate of change is, as we have said, decreasing in an absolute sense. That is,dQdt

starts out as quite negative but then goes to closer to zero. Then, dQdt

is actually

increasing, going from a more negative number to a less negative number. If the

quantity dQdt

is increasing, that means that

d

dt

(dQ

dt

)> 0

The derivative of the derivative is the second derivative, so we see that

d2Q

dt2> 0

To prove this more formally, we can differentiate the equation

dQ

dt= −rQ(t)

with respect to t. Since this equation holds true at all values of t, we can differentiate

it:d

dt

(dQ

dt

)=

d

dt(−rQ(t))

which means thatd2Q

dt2= −rdQ

dt

since r is a constant. But we already know that

dQ

dt= −rQ(t)

sod2Q

dt2= −r (−rQ(t)) = r2Q(t)

which is greater than zero since Q(t) is greater than zero. So

Answer:d2Q

dt2= r2Q(t) > 0


The particular differential equation F = ma:

28. The force of gravity on an object of mass m is −mg, where g is Earth’s gravitational

constant in meters per second squared and the negative sign is because the object

is “falling.” What is the acceleration due to gravity on an object?

Newton’s second law indicates that

F = ma

The force exerted on an object is equal to the product of the object’s mass and the

object’s acceleration. The force of gravity on an object can be measured as

F = −mg

so by Newton’s second law we have

ma = −mg

or

Answer:d2x

dt2= a = −g

The object that is in free fall continues to accelerate, or grow in downwards speed,

at the rate g which is constant for the Earth. On Earth, g is approximately 9.8 meters

per second squared. If the object starts at zero velocity, after one second of fall

the object’s speed will then be 9.8 meters per second downwards. If the object is

actually on Earth’s surface, the acceleration due to gravity does not produce any

velocity because it is cancelled by the upwards-pointing force of the Earth’s surface.

If there is no upwards-pointing force, however, the object in free fall will experience an

acceleration equal to −g regardless of its weight. The famous conclusion of Newton

is that (in the absence of air resistance) objects of different weights will fall according

to the same trajectory.


29. What is the velocity at time t of an object moving only under the influence of gravity, if

its initial velocity is zero? Is the velocity dependent on the initial height? Why or why

not?

The constant acceleration of an object moving only under the influence of gravity is

d2x

dt2= −g

This is a basic differential equation that is part of basic physics. We consider what

functions x(t) will satisfy this equation. Actually, for this problem, we are only required

to find dxdt

. We integrate both sides of the equation with respect to t.1 We have∫ (d

dt

(dx

dt

))dt =

∫−gdt

By the fundamental theorem of calculus, this equation is

dx

dt= −gt+ C

where C is the constant of integration. The velocity dxdt

is a function of t and a solution

to the differential equation. More precisely, it solves the differential equation

dv

dt= −g

tying the change in velocity to time. When t = 0 the velocity is zero, so we have

C = 0

and the velocity is

Answer: v(t) =dx

dt= −gt

Only this velocity will produce a constant acceleration of magnitude g and also satisfy

the initial condition that v0 = 0. The velocity is not dependent on the initial height,

because the velocity and acceleration due to gravity are not in any way dependent on

the initial height.21More precisely, we integrate from t = 0 to a time T .2This is for the purposes of this problem, which assumes no differences in gravity with height and also no differences in air resistance

at different heights.


30. What is the velocity at time t of an object moving only under the influence of gravity,

if its initial velocity is v0? Is the velocity dependent on the initial height? Why or why

not? Do the units check out in this equation for velocity?

The general solution for velocity in the equation

d2x

dt2=dv

dt= −g

is

v(t) = −gt+ C

Here, we have

v(0) = v0

so

C = v0

and the velocity is given by Answer: v(t) = −gt+ v0 . This is the case if the object

in free fall starts with some initial velocity, either up (if the object is tossed upwards)

or down (if the object is hurled downwards). Once again, velocity and acceleration

are independent of the initial height. The units do match in this equation. The units

of v(t) arem

s

The units of −gt are also ms

since g is in ms2

so

gt =m

s2· s = m

s

The units of v0 are also ms

, so this equation makes physical sense as well as mathe-

matical sense.



if the initial height is h and the initial velocity is zero?

The object’s acceleration is a constant

d2x

dt2= −g

This is a differential equation that we can solve for x(t). We have found that

dx

dt= v(t) = −gt+ v0

The initial velocity is zero, so in this problem,

dx

dt= −gt

Then, integrating again, we find that

x(t) =

∫−gtdt = −g

2t2 + C

Since

x(0) = h

we have Answer: x(t) = −g2t2 + h . This is the position of an object dropped from

a height h, and the unique specific solution to the differential equation d2xdt2

= −g with

the given initial conditions.



if the initial height is h and the initial velocity is v0? Is the position dependent on the

initial velocity? Why or why not? Do the units check out in this equation for position?

We have solved the differential equation that relates an object’s acceleration to the

force of gravity, to see that

v(t) = −gt+ v0

Now we seek to find the position of this object. We have

x(t) =

∫v(t)dt =

∫−gt+ v0dt = −

g

2t2 + v0t+ C

where C is a constant of integration. Since

x(0) = h

we have that

h = −g2(0)2 + v0(0) + C

so

C = h

and the position is given by Answer: x(t) = −g2t2 + v0t+ h . Here, the position is

dependent on the initial velocity, since the initial velocity continues to affect the tra-

jectory even though −g2t2 is the dominant term for large values of t. The units for this

equation do check out. The position x(t) is measured in meters. All three terms on

the right-hand side are measured in meters. We have that

−g2t2 =

m

s2· s2 = m

Likewise,

v0t =m

s· s = m

and

h = m


so this equation makes physical sense.


33. An object falls out of an helicopter that is flying at 1000m altitude. How long does

it take the object to reach the ground? Assume that g = 9.8ms2

. Also assume no air

resistance or other effects.

We know that the trajectory of the object is

x(t) = −g2t2 + h

since the initial velocity is zero (given that the object falls freely). The initial height is

1000m. We want to find the time when x(t) = 0 - that is, the time t when

0 = −g2t2 + 1000

This is

t =

√2000

g≈ Answer: t ≈ 14.3s

The units of 2000g

is s2, since 2000 is in meters and g is in ms2

. Then, the square root

yields a quantity in seconds, as is proper for time. It takes a little less than 15 seconds

for this object to fall a kilometer.


34. An object is thrown out of a helicopter that is flying at 500m altitude. The object is

thrown upwards at 10ms

. How long does it take the object to reach the ground? As-

sume that g = 9.8ms2

. Also assume no air resistance or other effects.

The trajectory of this object is

x(t) = −g2t2 + v0t+ h = −g

2t2 + 10t+ 500

since v0 = 10, h = 500. We want to solve for when x(t) = 0: that is,

−g2t2 + 10t+ 500 = 0

The solution to this is given by the quadratic formula:

t =−b±

√b2 − 4ac

2a=−10±

√100 + 1000g

−g=

10± 10√1 + 10g

g

We take the positive solution since the negative solution makes no sense in the con-

text of this problem (before t = 0, gravity is not the only force operating on the object;

to determine at what time before the drop time the helicopter lifted the object off the

ground at height 0, we would need to know about the speed of the helicopter). So

t =10± 10

√1 + 10g

g

which calculates to

Answer: t ≈ 11.2s

using g = 9.8ms2

. We can verify that the units for t are indeed seconds since the

denominator is in ms2

and the numerator is in ms

. The object takes a little longer than

11 seconds to reach the ground in this scenario.


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