1. If y′= x(1+ y) and y > -1 , then y=

Question

A) y=sinx
B) y=3x^2+C <<
C) y=Ce^1/2x^2-1
D) y=1/2e^x^2+C
E) y=Csq(x+3)

2. If dy/dx = x cos x^2 and y = −3 when x = 0, when x = π, y = ___.
A) -3.215
B) √2 <<
C) 1.647
D) 6
E) 3π

3. Suppose an experimental population of amoeba increases according to the law of exponential growth. There were 100 amoeba after the second day of the experiment and 300 amoeba after the fourth day. Approximately how many amoeba were in the original sample?
A) 5
B) 33 <<
C) 71
D) 10
E) Not enough information to determine

4. The velocity of a particle on the x-axis is given by the differential equation. dx/dt=3/4t^1/2
The particle is at x = 5, when t = 1. The position of the particle as a function of time is x(t) =

A)x(t)=1/8+5 <<<
B)x(t)=-1/2t+5
C)x(t)=1/2t-5
D)x(t)=-1/2t^3/2+11/2
E)x(t)=e^t+5

Answer 1

#1 ???

#2.
dy/dx = x cos x^2
y = (1/2) sin x^2 + C
when x=0, y = -3
-3 = (1/2) sin 0 + c
-3 = 0 + c
c = -3
so y = (1/2)sin x^2 - 3
when x = π
y = (1/2) sin (π^2) - 3
= -3.215...

#3.
let N = a e^(kt)

when t = 2 ---> 100 = a e^(2k)
when t = 4 ---> 300 = a e^(4k)
divide them

300/100 = a e^(4k)/( a e^(2k))
3 = e^(2k)
2k = ln3
k = ln3/2

so N = a e^( (ln3/2)t)
when t = 2
100 = a e^((ln3/2)(2) )
100 = a e^(ln3)
100 = a (3)
a = 100/3= appr 33

You are correct.

#4

if dx/dt = (3/4) t^(1/2)
then
x = (1/2) t^(3/2) + c
when x = 5, t = 1
5 = (1/2)(1^(3/2)) + c
5 = 1/2 + c
4.5 = c

x = (1/2) t^(3/2) + 4.5

check:
when t = 1, x = (1/2) (1^(3/2)) + 4.5
= 1/2 + 4.5 = 5

dx/dt = (3/2)(1/2) t^(1/2) = (3/4) t^(1/2)

mmmhhh?

Answer 2

For #1,
y′= x(1+ y)
write
y'/(1+y) = x
put p=y+1
p'/p = x
ln(p) = x²/2 + C
p=e^x²/2+C
y=e^x²/2+C-1
=Ce^x²/2-1 or
=Ce^x²/2

Answer 3

Ce^x^2/2 is close on #1, the correct answer is Ce^1/2x^2 -1
-3.215 is correct for #2
33 is correct for #3
3/4 t^1/2 is INCORRECT for #4 (so is 3/8 t^-1/2 + 5, another common multiple choice answer, though I'm not sure the correct answer here. Process of elimination though!)