Differential equations

Question 1

derivative of e^ax

Answer 1

ae^ax

Question 2

∫e^axdx

Answer 2

(1/a)e^ax+c

Question 3

∫(1/(ax+b))dx=

Answer 3

(1/a)ln|ax+b| +c

Question 4

∫(f’(x)/f(x))dx

Answer 4

(1/a)tan^-1(x/a)+c

Question 5

∫(1/sqrt(a²-x²))dx

Answer 5

sin^-1(x/a) +c

Question 6

forming a differential equation

Answer 6

if rate of growth∝size of the population (P), time (t)

rate of growth per hour is dP/dt

dP/dt ∝ P or dP/dt =kp where k

Question 7

growth of population P at time t is given by 0.05P²

write as a differential equation

Answer 7

dP/dt =0.05P²

0.05 being k

Question 8

temperature T^oC, t minutes

increasing rate (2t+0.01t2)

write as a differential equation

Answer 8

dT/dt=2t+0.01t²

Question 9

population P, t years, decreasing rate (0.1Pt)

Answer 9

dP/dt = -0.1Pt

Question 10

s metres, t seconds, s decreasing rate (0.02s) metres per second

write a differential equation

Answer 10

ds/dt = -0.02s

Question 11

t weeks after launch, N total number sold, rate of total growth N(20000-n)

a) write a differential equation including a constant k
b) explain why the model predicts that the total number sold will not exceed 20000

Answer 11

a) dN/dt=kN(20000-N)
b) increase stops wwhen dN/dt =0

if N>20000 then dN/dt <0 which is impossible

Question 12

solving by separating variables

Answer 12

finding an equation connecing the two variables, for instance P and t

Question 13

solve dP/dt =0.1P by separating variables

Answer 13

dP=0.1Pdt

1/pdP=0.1dt

∫(1/P)dP=∫0.1dt

ln|P|=0.1t+c

e^ln|P|=e^0.1t+c

|P|=Ae^0.1t (where A =e^c)

P=Ae^0.1t

Question 14

steps to solve a differential equation by variable separation

Answer 14

rearange so both sides only have one variable each

integrate both sides

use information about known values to fix the constant

Question 15

show dy/dx=2x(y+4), y>0. its solution is y=Ae^{x^2}-4

Answer 15

dy/y+4 =2xdx

∫(1/y+4)dy = ∫2xdx

ln(y+4)=x²+c since y>0 |y+4| is unnecessary

y+4=e^{x^2+x}=e^{x^2}e^c=Ae^{x^2}(where A=e^c)

y=Ae^{x^2}-4

Question 16

depth increase ∝ square root of depth

initial depth= 4m

after t houres depth =hm

a) write a differential equation
b) show sqrt(h)= (1/2)kt +2

Answer 16

a) dh/dt ∝ sqrt(h)

dh/dt=ksqrt(h)

b)show sqrt(h) =1/2kt+2

dh/sqrt(h) =kdt

h^-1/2dh=kdt

∫h^-1/2dh=∫kdt

2h^1/2=kt+c

sqrt(h)=1/2(kt+c)

sqrrt(4)=1/2(kx0 +c)

2=(1/2)c

c=4

sqrt(h)=1/2(kt+4) =(1/2)kt+2

Question 17

exponential growth and decay

population modelled by P=r-5e^-0.5t r,s are constants, t= years.

a) P=200 when t=0, P=360 when t=4 find r and s
b) what happens to the population as t gets larger and larger
c) at what rate is the population growing when t=6

Answer 17

a) r-s=200

r-0.135s=360

200+s=0.135s+360

s-0.135s=160

s=185

r-185=200

r=385

b) P=385-185e^-0.5t

t—>∞ P gets closer to 385

c)dP/dt=-185x(-0.5e^-0.5t)=92.5e^-3 =4.61 per year

Question 18

P=Ax(1.08)^t A constant, t= years

a) Find P when the population was first counted
b) How many years after the first count will the population reach 10000
c) what is the rate at which population is increasing 3 years after the first ount

Answer 18

P=5400, t=3

will be A

a) A= 5400/1.08³ = 4290
b) 10000=4290x1.08^t
tln0. 8=ln(10000/4290)

t=11.0

c) P=4290x1.08^t

dP/dt =4290x ln1.08 x 1.08^t

t=3, dP/dt=416 individuals a year

Question 19

a curve with parametric equation x=t + 1/t, y= t- 1/t, t>0

a) find the tangent where t=2, write in form ax+by=c
b) find the equation of the normal where t=2

Answer 19

dx/dt = 1- 1/t², dy/dt =1+ 1/t²

dy/dx=dy/dt/dx/dt =1+(1/t²)/1-(1-t²)

when t=2, dy/dx =5/3

when t=2 the coordinates are (2+(1/2),2-(1/2)) or (5/2,3/2)

gradient of tangent is 5/3 so the equation

y- (3/2) =(5/3)(x-(5/2))

5x-3y=8

b)gradient of normal = -3/5

y-(3/2)=-(3/5)(x-(5/2))

3x+5y=15

Question 20

differentiate xy+y=9 with respect to x

Answer 20

(x)dy/dx+y+dy/dx=0

Question 21

differentiate 2x²+xy+y²=28

find equations of tangent and normal at (3,2)

Answer 21

dy/dx=4x +((x)dy/dx +y) +(2y)dy/dx =0

x=3 and y=2

4(3) +(3dy/dx +2) +4dy/dx =0

dy/dx(3+4)=-14

dy/dx=-2

tangent gradient at (3,2) =-2

2x+y-8=0

gradient of normal at (3,2) is 1/2

x-2y+1=0

Question 22

dy/dx of y²

Answer 22

2y(dy/dx)

Question 23

exponential growth is in the form

Answer 23

y=ae^bt

Question 24

exponential decay is in the form

Answer 24

y=ae^-bt

Question 25

y gets closer and closer to c as t —->∞

Answer 25

y=c + ae^-bt

Question 26

what is equivalent to the expresssion a^x and what is its derivative

Answer 26

e^(lna)x its derivative is (lna)a^x