Differential Equations

There are many situations in science and business in which variables increase or decrease at a certain rate.

A differential equation expresses the rate at which one quantity varies in relation to another.

If the differential equation is solved then a direct relationship between the two variables can be found.

Graphing Differential Equations

The expression tells us that the graph of

2x

if x = 1 then the gradient is if x = 2 then the gradient isif x = 3 then the gradient is

How can this be shown on a diagram?

dy2x

dx

y has a gradient of

2

64

Each of the red lines has a gradient equal to twice the x coordinate

If the points are joined a family of curves results

Slope Field Graph

–6 –4 –2 2 4 6

–4

–3

–2

–1

1

2

3

4

x

y

–6 –4 –2 2 4 6

–4

–3

–2

–1

1

2

3

4

x

y

At every point on the curve the gradient is equal to twice the x coordinate

dy2x

dx ie the gradient equals twice the x coordinate

-1

-2

-3

-4

1

2

3

4

-1-2-3-4-5 1 2 3 4 5 60X->

|̂Y

If the gradient is x then to find y we :

integrate

as integration and differentiation are the :

reverse of each other

= x2 + C

From the slope field graph it can be seen that all the graphs are vertical translations of each other.

So C represents a vertical translation

2xdxy =

Graphing Differential Equations

The expression tells us that the graph of

y

if y = 1 then the gradient is if y = 2 then the gradient isif y = 3 then the gradient is

How can this be shown on a diagram?

y has a gradient of

1

32

dyy

dx

–6 –4 –2 2

–1

1

2

3

4

5

6

x

y

-1

1

2

3

4

5

6

-1-2-3-4-5-6 1 20X->

|̂Y

–6 –4 –2 2

–1

1

2

3

4

5

6

x

y

At every point on the curve the gradient is equal to the y coordinate

Each of the red lines has a gradient equal to the y coordinate

Slope Field Graph

If the points are joined a family of curves results

dyy

dx ie the gradient equals the y coordinate

If the gradient is x then to find y we :

integrate

as integration and differentiation are the :

reverse of each other

So how do we integrate an expression where the gradient depends on y rather than x

dyy

dx

Reminder siny = ex + C

This must be differentiated using implicit differentiation•When differentiating y’s write dy•When differentiating x’s write dx•Divide by dx

Implicit Differentiation

Implicit Differentiation

For example siny = ex + C

cosy dy = ex dx

dy

dxRearrange to make the subject

xdy e

dx cos y

xdycos y e

dx Divide by dx

–6 –5 –4 –3 –2 –1 1 2

–4

–3

–2

–1

1

2

x

y

Slope Field Graphxdy e

dx cosy

Differentiating

siny = ex + C

cosy dy = ex dx

xdycos y e

dx

Reversing the Process

xdy e

dx cos y

xdycos y e

dx

cosy dy = ex dx

siny = ex + C

xcosy dy = e dx

mul

tiply

by

cosy

multiply by dxintegrate

Integrating

Finding the constant Csiny = ex + CTo find the constant C a boundary condition is needed.

If we are told that when x = 0 then y = /2 then we can find C.

siny = ex + C

Substitute x = 0 and y = /2

sin /2 = e0 + C

So C = 0 siny = ex

y = sin-1(ex)

y= sin-1(ex) when x = 0 then y = /2

–6 –5 –4 –3 –2 –1 1

–1

1

2

3

x

y

Slope Field Graph

You try this one

2

y

dy x

dx e when x = 3 y = 0

ey dy = x2 dx

y 2e dy = x dx 3

y xe c

3

3y x

lne ln( 8)3

3x

y ln( 8)3

when x = 3 y = 0 0e 9 c

c = -8

2

y

dy x

dx e

Slope Field Graph

when x = 3 y = 0

1 2 3 4 5 6 7 8

–4

–3

–2

–1

0

1

2

3

4

5

6

x

y3xy ln( 8)

3