Objective:

to form and solve firstorder differential equations

DIFFERENTIAL EQUATIONS

DIFFERENTIAL EQUATIONS

4dxdy

18 tdtdr

23 ttcosdtdv

Here are some first order differential equations.

A differential equation is simply an equation that contains one

or more derivatives.

te

dtdp p37

DIFFERENTIAL EQUATIONS

pedzdp

dzpd 42

2

34

1232 2

2

3

3

ds

wdds

wd

A second order differential equation.

A third order differential equation.

Exposure to Iodine-131 can cause hypothyroidism, thyroid nodules and

cancer.

Iodine-131 is a short-lived radioactive isotope.

It enters the air from nuclear power plants.

The quantity of Iodine-131 in the environment reduces by half every

8.1 days.

m = mass of Iodine-131 in grams t = time elapsed in days

t.em 08560

t.em 08560

t.em 08560

t.e.dtdm 0856008560

t.em 08560

t.e.dtdm 0856008560 m.

dtdm

08560

t.em 08560

m.dtdm

08560t.e.dtdm 0856008560

mkdtdm

The rate of decay of

Iodine-131 is proportional to its

current mass.

Express the following as differential equations:

The rate at which the population of rabbits increases is proportional to the number present.

Express the following as differential equations:

NkdtdN

The rate at which the population of rabbits increases is proportional to the number present.

Express the following as differential equations:

The rate at which the population of rabbits increases is proportional to the number present.

The volume of water leaking out of a tank is at a rate proportional to the volume of the water.

NkdtdN

Express the following as differential equations:

The rate at which the population of rabbits increases is proportional to the number present.

The volume of water leaking out of a tank is at a rate proportional to the volume of the water.

VkdtdV

NkdtdN

Express the following as differential equations:

The rate at which the population of rabbits increases is proportional to the number present.

The volume of water leaking out of a tank is at a rate proportional to the volume of the water.

The rate at which a tree grows is proportional to the time since it was planted.

VkdtdV

NkdtdN

Express the following as differential equations:

The rate at which the population of rabbits increases is proportional to the number present.

The volume of water leaking out of a tank is at a rate proportional to the volume of the water.

The rate at which a tree grows is proportional to the time since it was planted.

VkdtdV

tkdtdh

NkdtdN

Express the following as differential equations:

The rate at which the population of rabbits increases is proportional to the number present.

The volume of water leaking out of a tank is at a rate proportional to the volume of the water.

The rate at which a tree grows is proportional to the time since it was planted.

The temperature of an object is decreasing at a rate proportional to the difference between the temperature of the object and the background temperature of 20°C.

VkdtdV

tkdtdh

NkdtdN

Express the following as differential equations:

The rate at which the population of rabbits increases is proportional to the number present.

The volume of water leaking out of a tank is at a rate proportional to the volume of the water.

The rate at which a tree grows is proportional to the time since it was planted.

The temperature of an object is decreasing at a rate proportional to the difference between the temperature of the object and the background temperature of 20°C.

VkdtdV

tkdtdh

NkdtdN

)(kdtd

20

The rate of absorption of a drug by the kidneys is proportional to the amount of drug present. After t hours, the quantity of the

drug is x mg. 25 mg of the drug is administered to a patient and 10 mg remains 2 hours later. Write down a differential equation relating x and t and then use the information provided to solve it.

The rate of absorption of a drug by the kidneys is proportional to the amount of drug present. After t hours, the quantity of the

drug is x mg. 25 mg of the drug is administered to a patient and 10 mg remains 2 hours later. Write down a differential equation relating x and t and then use the information provided to solve it.

xkdtdx

The rate of absorption of a drug by the kidneys is proportional to the amount of drug present. After t hours, the quantity of the

drug is x mg. 25 mg of the drug is administered to a patient and 10 mg remains 2 hours later. Write down a differential equation relating x and t and then use the information provided to solve it.

xkdtdx

xdx

dtk

The rate of absorption of a drug by the kidneys is proportional to the amount of drug present. After t hours, the quantity of the

drug is x mg. 25 mg of the drug is administered to a patient and 10 mg remains 2 hours later. Write down a differential equation relating x and t and then use the information provided to solve it.

xkdtdx

xdx

dtk xdx

dtk

The rate of absorption of a drug by the kidneys is proportional to the amount of drug present. After t hours, the quantity of the

drug is x mg. 25 mg of the drug is administered to a patient and 10 mg remains 2 hours later. Write down a differential equation relating x and t and then use the information provided to solve it.

xkdtdx

xdx

dtk xdx

dtk cxlntk

The rate of absorption of a drug by the kidneys is proportional to the amount of drug present. After t hours, the quantity of the

drug is x mg. 25 mg of the drug is administered to a patient and 10 mg remains 2 hours later. Write down a differential equation relating x and t and then use the information provided to solve it.

xkdtdx

xdx

dtk xdx

dtk cxlntk

When t = 0, x = 25:

The rate of absorption of a drug by the kidneys is proportional to the amount of drug present. After t hours, the quantity of the

drug is x mg. 25 mg of the drug is administered to a patient and 10 mg remains 2 hours later. Write down a differential equation relating x and t and then use the information provided to solve it.

xkdtdx

xdx

dtk xdx

dtk cxlntk

When t = 0, x = 25: 25lnc

The rate of absorption of a drug by the kidneys is proportional to the amount of drug present. After t hours, the quantity of the

drug is x mg. 25 mg of the drug is administered to a patient and 10 mg remains 2 hours later. Write down a differential equation relating x and t and then use the information provided to solve it.

xkdtdx

xdx

dtk xdx

dtk cxlntk

When t = 0, x = 25: 25lnc 25x

lntk

The rate of absorption of a drug by the kidneys is proportional to the amount of drug present. After t hours, the quantity of the

drug is x mg. 25 mg of the drug is administered to a patient and 10 mg remains 2 hours later. Write down a differential equation relating x and t and then use the information provided to solve it.

xkdtdx

xdx

dtk xdx

dtk cxlntk

When t = 0, x = 25: 25lnc 25x

lntk

When t = 2, x = 10:

The rate of absorption of a drug by the kidneys is proportional to the amount of drug present. After t hours, the quantity of the

drug is x mg. 25 mg of the drug is administered to a patient and 10 mg remains 2 hours later. Write down a differential equation relating x and t and then use the information provided to solve it.

xkdtdx

xdx

dtk xdx

dtk cxlntk

When t = 0, x = 25: 25lnc 25x

lntk

When t = 2, x = 10: 45802

40.

.lnk

The rate of absorption of a drug by the kidneys is proportional to the amount of drug present. After t hours, the quantity of the

drug is x mg. 25 mg of the drug is administered to a patient and 10 mg remains 2 hours later. Write down a differential equation relating x and t and then use the information provided to solve it.

xkdtdx

xdx

dtk xdx

dtk cxlntk

When t = 0, x = 25: 25lnc 25x

lntk

When t = 2, x = 10: 45802

40.

.lnk

254580

xlnt.

The rate of absorption of a drug by the kidneys is proportional to the amount of drug present. After t hours, the quantity of the

drug is x mg. 25 mg of the drug is administered to a patient and 10 mg remains 2 hours later. Write down a differential equation relating x and t and then use the information provided to solve it.

xkdtdx

xdx

dtk xdx

dtk cxlntk

When t = 0, x = 25: 25lnc 25x

lntk

When t = 2, x = 10: 45802

40.

.lnk

254580

xlnt.

254580 x

e t.

The rate of absorption of a drug by the kidneys is proportional to the amount of drug present. After t hours, the quantity of the

drug is x mg. 25 mg of the drug is administered to a patient and 10 mg remains 2 hours later. Write down a differential equation relating x and t and then use the information provided to solve it.

xkdtdx

xdx

dtk xdx

dtk cxlntk

When t = 0, x = 25: 25lnc 25x

lntk

When t = 2, x = 10: 45802

40.

.lnk

254580

xlnt.

254580 x

e t.

The rate of absorption of a drug by the kidneys is proportional to the amount of drug present. After t hours, the quantity of the

drug is x mg. 25 mg of the drug is administered to a patient and 10 mg remains 2 hours later. Write down a differential equation relating x and t and then use the information provided to solve it.

xkdtdx

xdx

dtk xdx

dtk cxlntk

When t = 0, x = 25: 25lnc 25x

lntk

When t = 2, x = 10: 45802

40.

.lnk

254580

xlnt.

t.ex 458025 25

4580 xe t.

hdtdh

1620

A hot-air balloon can reach a maximum height of 1.25 km and the

rate at which it gains height decreases as it climbs according to the formula:

where h is the height in km and t is the time in hours since lift-off. How long does the balloon take to reach 1 km?

A hot-air balloon can reach a maximum height of 1.25 km and the

rate at which it gains height decreases as it climbs according to the formula:

where h is the height in km and t is the time in hours since lift-off. How long does the balloon take to reach 1 km?

hdh

dt1620

hdh

dt1620

hdtdh

1620

hdh

dt1620

hdh

dt1620

k)h(lnt 162016

1

When t =0 hours, h = 0 km so:

hdh

dt1620

k)h(lnt 162016

1

When t =0 hours, h = 0 km so: 2016

1lnk

hdh

dt1620

k)h(lnt 162016

1

When t =0 hours, h = 0 km so: 2016

1lnk

20162016

1

16

1ln)h(lnt

When h = 1 km:

hdh

dt1620

k)h(lnt 162016

1

When t =0 hours, h = 0 km so: 2016

1lnk

20162016

1

16

1ln)h(lnt

When h = 1 km: 20416 lnlnt

hdh

dt1620

k)h(lnt 162016

1

When t =0 hours, h = 0 km so: 2016

1lnk

20162016

1

16

1ln)h(lnt

20416 lnlnt When h = 1 km:

516 lnt

hdh

dt1620

k)h(lnt 162016

1

When t =0 hours, h = 0 km so: 2016

1lnk

20162016

1

16

1ln)h(lnt

20416 lnlnt When h = 1 km:

516 lnt hours.t 1010

The balloon takes 6 minutes to rise to 1 km

hdh

dt1620

k)h(lnt 162016

1

When t =0 hours, h = 0 km so: 2016

1lnk

20162016

1

16

1ln)h(lnt

20416 lnlnt When h = 1 km:

516 lnt hours.t 1010

The balloon takes 6 minutes to rise to 1 km