# It's only a minus sign

Solve these differential equations to see how a minus sign can
change the answer

## Problem

In this problem we shall see how a simple minus sign in a differential equation can completely change the character of the solution.

Two particles are released from $x = 1$ at time $0$ and their speed at any point x will be given by these two differential equations:

particle $A \quad \quad$ | $\frac{dx}{dt}=x$ |

particle $B$ | $\frac{dx}{dt}=-x$ |

Without solving the equations, can you describe how the
particles will move? Draw a sketch graph of the path you expect the
particles to take.

Now solve the equations to see if you were correct.

Next suppose that two more particles are released with a
positive velocity at time $0$ from the origin and move according to
these equations, in which v is the velocity of each particle:

particle $C \quad \quad$ | $\frac{dv}{dt}=x$ |

particle $D$ | $\frac{dv}{dt}=-x$ |

Without solving the equations, can you provide a clear
description of the subsequent motion of the particles?

Would releasing the particles with a negative velocity from the origin have a significant effect on the type of motion which results?

Could you find initial starting points and velocities which would give rise to motions in which the particles slow down and stop?

## Getting Started

Remember that the differential of x means the 'rate of change' of $x$. The equation tells us exactly what that rate of change must be at each point.

What does a positive rate of change tell us about the changes in $x$? What does a negative rate of change tell us about the changes in $x$?

## Student Solutions

Particle A

Qualitative behaviour: moves in positive direction with an
increasing velocity. $$\frac{dx}{dt} = x \Rightarrow \int
\frac{1}{x}dx = \int dt \Rightarrow x = Ae^t$$ $x(0) = 1$ so $A =
1$ and the solution is $x(t) = e^t$.

Particle B

Qualitative behaviour: moves in negative
direction with a decreasing velocity, gradually approching origin.
$$\frac{dx}{dt} = x \Rightarrow x = Ae^{-t}$$ $x(0) = 1$ so $A = 1$
and the solution is $x(t) = e^{-t}$.

Particle C

Qualitative behaviour: moves in
positive direction with an increasing velocity. $$\frac{dv}{dt} = x
\Rightarrow \frac{d^2x}{dt^2} - x = 0 \Rightarrow x = Ae^t +
Be^{-t}$$ If particle starts at origin with velocity $v$ then $x(t)
= \frac{v}{2}(e^t - e^{-t})$.

Particle D

Qualitative behaviour: oscillates
around origin. $$\frac{dv}{dt} = -x \Rightarrow \frac{d^2x}{dt^2} +
x = 0 \Rightarrow x = Ae^{it} + Be^{-it} = C\sin{t} + D\cos{t}$$ If
particle starts at origin with velocity $v$ then $x(t) =
v\sin{t}$.

Particles C and D have symmetrical
equations of motion, in the sense that starting at origin with
negative velocity would result in the same motion but in the
opposite direction.

Observe that if we start particle C
at $+1$ with velocity $-1$ then our equation of motion is $x(t) =
e^{-t}$, so the particle eventially stops at the origin. Are there
any other initial conditions that can be imposed on particles C or
D which force the particle to stop eventually?

## Teachers' Resources

The key for this problem is for students to appreciate that the signs in differential equations are of crucial importance in determining the structure of a solution. They can make the difference between solutions growing to infinity, oscillating or settling down to zero.

When constructing differential equations using, for example, $F \; = \; ma$, negative signs on the right hand side correspond to 'repulsions' and positive signs correspond to 'attractions'.

Understanding the structure of equations in this way is a very powerful approach which can transcend the details of the algebraic solution.