Question

These are all the questions I missed on my practice quizzes, however I was never given the correct answers. I was hoping someone could give me the answers to these so I'd be able to study them! (I know some of them were simple/dumb mistakes :'))

1. Which of the following is a separable, first-order differential equation?
dy/dx = (x+y)/(2x)
dy/dx = (x+y)/(x-y)
dy/dx = sinx
dy/dx = (xy)/(x+y)
dy/dx = 2Inxy

2. Which of the following graphs is consistent with the differential equation dy/dx = -(x/y)?
First and second graph: (
gyazo.com/0252da0870a88a38a1f4d37d691cdd9c)
third and fourth graph: (gyazo.com/6d1e0d4f19fd2a62e0fe2999d4bdc4b1)

3. The differential equation dy/dx = y/(y^2) has a solution given by:
y = C sqrt(x^2+4)
y = Ce^(-1/x)
y = (-1/2)e^(x^2)+C
x^2 + y^2 = C
y = Ce^(-kt)

4. The family of curves represented on this graph would appear to be consistent with which of the following differential equations?
Graph: gyazo.com/f580adf38cc523b7651053409c0acf5a
dy/dx = y/x
dy/dx = tan(x+y)
dy/dx = x sqrt(x-3)
dy/dx = 3x^2
dy/dx = x+y

5. yy' - e^x = 0 and y=4 when x=0, which means that:
y=x-Inx^2 + 4
y^2 = 4x^2+3
y = -2x + (1/2)x^3
y^2 = 2e^x + 14
(1/2)In|x^2 + 4| + 6

6. The rate of change of P with respect to t is directly proportional to (10-t) and inversely proportional to the cube root of P. The differential equation that models this situation is:
dP/dt = (k(10-t))/((1)/(cube root(p))
dP/dt = (10-t)/(sqrt(p))
dP/dt = k (10-t) cube root(p)
dP/dt = (k(cube root(p))/(10-t)
dP/dt = (k(10-t))/(cube root(p))

7. If y' = x(1+y) and y>-1, then y=___.
y = sinx
y = 3x^2 + C
y = Ce^(x^2/2) - 1
y = (1/2)e^(x^2) + C
y = C sqrt(x+3)

8. If dy/dx = xcosx^2 and y = -3 when x=0, when x=pi, y=___.
-3.215
sqrt2
1.647
6
3pi

9. 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?
5
33
71
10
Not enough info to determine.

10. The velocity of a particle on the x-axis is given by the differential equation dx/dt - (3/4)t^(1/2).
The particle is at x=5, when t=1. The position of the particle as a function of time is x(t) = ___.
x(t) = (1/8)t + 5
x(t) = -(1/2)t + 5
x(t) = (1/2)t - 5
x(t) = (1/2)t^(3/2) + (9/2)
x(t) = e^t + 5

Answer 1

#1 dy/dx = F(x,y)
F(x,y) can be factored into f(x)*g(y)
that is, the variables can be separated.
So, C, where g(y) = 1

#2 I didn't check the graphs, but recall that if you have
x^2+y^2 = r^2
2x + 2y y' = 0
y' = -x/y

#3 dy/dx = y/(y^2) = 1/y
y dy = dx
1/2 y^2 = x+c
y^2 = 2x+c
y = √(2x+c)
I suspect a typo, since
dy/dx = y/x^2
dy/y = dx/x^2
lny = -1/x + ln c
y = c*e^(-1/x)

#4 Note that the slopes are zero along the line y = -x
So, expect dy/dx to be something involving x+y
Since tan(x+y) is periodic, and the slope field is not, my choice is
dy/dx = x+y

#5 yy' - e^x = 0 and y=4 when x=0
yy' = e^x
y dy = e^x dx
1/2 y^2 = e^x + c
y^2 = 2e^x + c
(0,4) ==>
16 = 2+c
y^2 = 2e^x + 14

#6 The rate of change of P with respect to t: dP/dt
is directly proportional to (10-t): k(10-t)
and inversely proportional to the cube root of P: 1/∛P
dP/dt = k(10-t)/∛P

#7 y' = x(1+y)
dy/(1+y) = x dx
ln(1+y) = 1/2 x^2 + lnc
1+y = c e^(1/2 x^2)
y = c e^(x^2/2) - 1

#8 dy/dx = xcosx^2
let u = x^2
dy/dx = 1/2 cosu
y = 1/2 sinu = 1/2 sin(x^2) + c
(0,-3) ==> -3 = 0 + c
y = 1/2 sin(x^2) - 3
y(π) = 1/2 sin(π^2) - 3 = -3.215

#9 y = ce^(kt)
(2,100) ==> ce^(2k) = 100
(4,300) ==> ce^(4k) = 300
now divide, and you get
e^(2k) = 3
k = ln3/2 = 0.549
ce^(1.099) = 100
c = 100/3
so, originally there were about 33

#10 dx/dt = (3/4)t^(1/2)
x = (3/4)(2/3)t^(3/2) = 1/2 t^(3/2) + C
(1,5) ==> 1/2 + C = 5
C = 9/2
x = 1/2 t^(3/2) + 9/2

Answer 2

Thank you so much for your help and even showing me the work!! I see what I got wrong on these now!

Answer 3

lol this is an apex test. cheater lmfao