Differential Equations Problem Set

Differential Equations Problem Set

University

10 Qs

quiz-placeholder

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Differential Equations Problem Set

Differential Equations Problem Set

Assessment

Quiz

Mathematics

University

Practice Problem

Medium

CCSS
HSF.LE.A.1, HSA.CED.A.2, HSF.LE.A.4

+4

Standards-aligned

Created by

jhayr ebreo

Used 29+ times

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10 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

The population of the little town of Scorpion Gulch is now 1000 people. The population is presently growing at about 5% per year.

Find the general solution.

P=.05e1000t

P=1000e5t

P=1000e.05t

p=lne1000t

Tags

CCSS.HSF.BF.A.1

CCSS.HSF.LE.A.1

CCSS.HSF.LE.A.2

CCSS.HSF.LE.B.5

2.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

Given the differential equation dP/Dt=5P

A) Find the general solution for P to the differential equation

B)Find the particular solution for P to the differential equation given P(0)= 418

A) P=Ce5t

B) P= 5e418t

A) P=Ce5t

B) P= 418e5t

A) P=Cet

B) P= 418et

A) P=Ce10t

B) P= 418e10t

Tags

CCSS.HSF.LE.A.1

CCSS.HSF.LE.A.2

CCSS.HSF.LE.B.5

3.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

y varies jointly with x and the square of z

y= kxz2

y= kx√z

y= kxz

kx2z

Tags

CCSS.HSA.CED.A.2

CCSS.HSA.SSE.A.1

4.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

y is proportional to the product of z and the square root of x

y= z(x2)

y= kz(x2)

y= kx√z

y=zkx

5.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

Write a differential equation for the statement:

"The rate of change of P is proportional to the product of P and 4 - P."

dP/dt = kP

dP/dt = k(4 - P)

dP/dt = kP(4-P)

dP/dt = kP/(4-P)

6.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

The number of bacteria increases at a rate proportional to the present amount. After 5 hours, there were 80 bacteria and after 8 hours, there were 120 bacteria. How many bacteria were present initially?

1

About 16

About 40

About 53

Tags

CCSS.HSF.LE.A.4

7.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

Greenium has a half-life of 23 years and decays following the Law of Exponential Change. After how many years will 80% of an original amount have decayed?

t = 7.4143 years

t = 10.2646

t = 53.4043 years

t = 74.0341 years

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