#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(Ï€) = 1/2 sin(Ï€^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