Question

Always reward medals, dy/dx=ysin(2x) when x=0 y=3

Answer 1

dy/dx=y*sin2x
\[\int\limits_{}^{} dy/y=\int\limits_{}^{}\sin2x*dx\]

Answer 2

lny=(-cos2x)/2+c
now at x=0 y=3
ln3=(-cos0)/2+c
c=ln3 +1/2
lny=-cox2x/2+ln3+1/2
lny-ln3=-cos2x/2+1/2=(1-cos2x)/2
ln(y/3)=(1-cos2x)/2
y/3=e^((1-cos2x)/2)
y=(1/3)*e^((1-cos2x)/2)

Answer 3

@hhuumm

Answer 4

answer key states y=3e^[sin^2(x)]

Answer 5

how did you get cos2x/2? integral of sin2x= -cos2x/2?

Answer 6

@gorv

Answer 7

yep

Answer 8

ok stick around just a little bit longer so i can wrap my head around it...

Answer 9

pls hurry

Answer 10

lol trust me my exam is in 40 minutes i'm moving

Answer 11

instead of adding ln and blah blah could i have just put both sides as an exponent to the E and called it a day (step 4)

Answer 12

try to put it u will get ur ans

Answer 13

despite the awk glamour shot avi, you are the man. Earned your medal thanks.