Question

How to solve (dy/dx)-(e^x) (sinx) =0? Separation of variables and integration.

Answer 1

just solve it ... dy/dx = e^x sin(x)
dy = e^x sin(x) dx
integrate both sides

Answer 2

Wait hold on srry that exponent of e is "y" not "x"

Answer 3

I integrated (1/e^y)dy=sinxdx, but then...is that -u^2/2=-ln(cosx)+c???

Answer 4

and so e^-2y=2ln|cosx|+c

Answer 5

and then I came to e^-2y=ln(cosx^2) +c and I don't really know how to deal with this...

Answer 6

\(\large \int \dfrac{1}{e^y}dy=\int e^{-y}dy = ....?\)
try this integral again ?

Answer 7

thats left side integral
for the right side, its simply
\(\Large \int \sin x dx\)

Answer 8

and the integral of sinxdx is -ln(cosx)+c

Answer 9

\(\Large \int \sin x dx = -\cos x +c\)

Answer 10

ohhh maybe I was thinking of tangetn hahah uhm... and the e^-y, use u -substitution?

Answer 11

i am not sure if you're allowed to use this directly,
\(\large \int \dfrac{1}{e^y}dy=\int e^{-y}dy =
\dfrac{e^{-y}}{-1}+c\)
dividing by the co-efficient of the variable

Answer 12

???

Answer 13

you didn't get that ?
then use a u-substitution as
-y = u

Answer 14

ohhhhhhh!

Answer 15

so then -e^-y=cosx+c, then ln both sides?

Answer 16

you need to isolate 'y' ? then yes!

Answer 17

uhm... where does the negative on the e go when we put the ln on?

Answer 18

first isolate e^(-y)
(note that right side is actually -cos x + c and not just cos x +c)

so, e^(-y) = cos x - c

now take ln on both sides

Answer 19

oh! right! I completely forgot the negative on the cosine... so then y=ln(cosx)^-1 +c?

Answer 20

lets see
\(\large \ln (e^{-y}) = \ln (\cos x -c)\)

Answer 21

\(\large -y \ln e = \ln (\cos x -c)\)

Answer 22

and since ln e = 1

Answer 23

\(\large y = -\ln (\cos x -c )\)
thats it!

Answer 24

yay! okay cool~ ^^ thank you!

Answer 25

welcome ^_^