Question

dy/dx = (cos x) e^(y+sinx) and y = 0 when x = 0. find the original equation

Answer 1

Original equation as in the integral of that function?

Answer 2

as an answer I get e^sinx + ln1 -1 =y which is not the right answer, according to the book

Answer 3

yes sir, we are to integrate via separation of variables

Answer 4

This website can help faster than I can. http://tutorial.math.lamar.edu/Classes/DE/SeparationofVariables.aspx

Where I went with all of my questions.

Answer 5

maybe you can check my work

\[(\cos x)(e ^{y+\sin x})\]
\[ \frac{ dy }{ e^y } = (\cos x) e^y dx\]
u = sinx and du = cos x
\[\ln e^y + c = e^{sinx} +c\]
find c by plugging in 0 and I get (ln 1) -1 = c
put it all together and I get e^[sinx] + ln 1 - 1 = y

Answer 6

First you must separate the variables.
\[\frac{ dy }{ dx } = cosx e ^{y+sinx}\]
\[\frac{ dy }{ dx }=cosx e ^{y}e ^{sinx}\]
\[\int\limits_{?}^{?}\frac{ dy }{ e ^{y} }=\int\limits_{?}^{?}cosxe ^{sinx}\]

for the left part of the equation, let u=sinx then du=cosxdx

I think you can do the rest. Cheers

Answer 7

actually, its after that where i begin to get stuck, can you go a little further? no need to use the equations button, just a verbal explanation will suffice

Answer 8

For the right side, let u = sinx, then du = cosxdx

Answer 9

so substitute u and du, this will result to the equation
\[\int\limits_{?}^{?}=e ^{u}du\]
cos x will be cancelled, after the integration, you will com up to \[e ^{u}\] wich is equal to \[e ^{sinx}\]

Answer 10

ln(e^y)+c=e^sinx+c

what happens after this?

Answer 11

integral of 1/ e^y is ln e^y isn't it?

Answer 12

On the left side, substitute u =-y and then -du=dy

Answer 13

ya, I was wrong. you got it, just substitute the constraints to find C

Answer 14

1/e^y = e^-y

Answer 15

holy cow, thats what i needed to see haha

Answer 16

wait, I think I was correct

Answer 17

Then substitute. Giving you the (negative) integral e^u du

Answer 18

yeah jeffrey i think you were right, I just did not get that you turned 1/e^y to e^-y as wade just showed me

Answer 19

ya, so therefore the final equation would be
\[-\frac{ 1 }{ e ^{y} }=e ^{sinx} +C\]

Answer 20

Yes....

Answer 21

There you go, substitute the constraints to find C.

Answer 22

alright sweet, wish i could give you both best answer, seriously, thanks to both of you

Answer 23

No prob dude! :)