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DE's and Mathematical Modelling Quiz 1

Total questions: 10

Worksheet time: 45mins

Name
Class
Date
1.

Shown above is a slope field for which differential equation?

a)

dydx= xy\frac{dy}{dx}=\ xy

b)

dydx=xy + y\frac{\text{d}y}{\text{d}x}=xy\ +\ y

c)

dydx=xy−y\frac{\text{d}y}{\text{d}x}=xy-y

d)

dydx=xy+x\frac{\text{d}y}{\text{d}x}=xy+x

e)

dydx=(x+1)3\frac{\text{d}y}{\text{d}x}=\left(x+1\right)^3

2.

∫(x−1)x dx =\int_{ }^{ }\left(x-1\right)\sqrt{x}\ dx\ =  

a)

12x2+2x32−x+c\frac{1}{2}x^2+2x^{\frac{3}{2}}-x+c  

b)

23x32+12x12+c\frac{2}{3}x^{\frac{3}{2}}+\frac{1}{2}x^{\frac{1}{2}}+c  

c)

32x−1x+c\frac{3}{2}\sqrt{x}-\frac{1}{\sqrt{x}}+c  

d)

25x52−23x32+c\frac{2}{5}x^{\frac{5}{2}}-\frac{2}{3}x^{\frac{3}{2}}+c  

e)

12x2−x+c\frac{1}{2}x^2-x+c  

3.

A rumour spreads among a population of N people at a rate proportional to the product of number of people who have heard the rumour and the number of people who have not heard the rumour. If p denotes the number of people who have heard the rumour, which of the following differential equations could be used to model this situation with respect to time t, where k is a positive constant?

a)

dpdt=kp\frac{dp}{dt}=kp

b)

dpdt=kp(N−p)\frac{dp}{dt}=kp\left(N-p\right)

c)

dpdt=kp(p−N)\frac{dp}{dt}=kp\left(p-N\right)

d)

dpdt=kt(N−t)\frac{dp}{dt}=kt\left(N-t\right)

e)

dpdt=kt(t−N)\frac{dp}{dt}=kt\left(t-N\right)

4.

Let y = f(t) be a solution to the differential equation  dydt=ky\frac{dy}{dt}=ky  , where k is a constant.  Values of f for selected values of t are given in the table above.  Which of the following is an expression for f(t)?

a)

4t + 4

b)

 2t2+42t^2+4  

c)

 et2ln⁡9+3e^{\frac{t}{2}\ln9}+3  

d)

 4et2ln⁡34e^{\frac{t}{2}\ln3}  

5.

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.

a)

P=.05e1000t

b)

P=1000e5t

c)

P=1000e.05t

d)

p=lne1000t

6.

2. Solve the differential equation
 dydt=3t2y\frac{dy}{dt}=\frac{3t^2}{y}  with initial condition y(2) = 0.

a)

𝑦 = ln(15t)

b)

𝑦 =  16t316t^3  

c)

 y=(2t3−16)12y=\left(2t^3-16\right)^{\frac{1}{2}}  

d)

𝑦 =  (2t3−16)\left(2t^3-16\right)  

7.

∫ 2x cos(x2) dx

a)

sin( x2x^2 ) + C

b)

2 sin ( 2x ) + C

c)

12\frac{1}{2} sin( x2x^2 ) + C

d)

4 cos(2x ) + C

8.
a)
ln(2x2+6) + C
b)
ln(2x2+6)(4x) + C
c)
1/(2x2+6) + C
d)
1/(2x2+6)2 + C
9.
∫(3x⁴+2)⁴(12x³)dx
a)
½(3x⁴+2)⁴+C
b)
(3x⁴+2)⁴+C
c)
⅕(3x⁴+2)⁵+C
d)
⅓(3x⁴+2)⁶+C
10.

 ∫100x35x4+3dx\int100x^3\sqrt{5x^4+3}dx  

a)

 152(5x4+3)3+C\frac{15}{2}\sqrt{\left(5x^4+3\right)^3}+C  

b)

 5(5x4+3)3+C5\sqrt{\left(5x^4+3\right)^3}+C  

c)

 (5x4+3)3+C\sqrt{\left(5x^4+3\right)^3}+C  

d)

 103(5x4+3)3+C\frac{10}{3}\sqrt{\left(5x^4+3\right)^3}+C