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use separation of variables to solve the initial value problem. dy/dx=(cos x)e^(y+sinx) and y=0 when x=0

Mathematics
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\[e^{-y} dy =\cos(x)e^{\sin(x)} dx\]
integrate both sides
Lol.. I don't know how to move "the x with dx" and "y with dy".. :/ can you show me that part?

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Other answers:

i multiplied dx on both sides
\[e^{y+\sin(x)}=e^{y} e^{\sin(x)}\]
divide both sides by e^y
1/e^{y}=e^{-y}
Oh ok. Thanks.
How do I integrate cosxe^(sin x) dx?
ley u=sin(x) => du=cos(x) dx
? What about the e? Do I put e to the u power?
\[\int\limits_{}^{}e^u du=e^u+C=e^{\sin(x)}+C\]
i replace sin(x) with u and cos(x) dx with du
Oh? But don't I integrate |dw:1327549907891:dw|? What happens with the cos...
cos(x) dx=du i replaced cos(x) dx with du i did integrate
|dw:1327550585485:dw|
|dw:1327550603285:dw|
Ohh I see. Ok, what do I do now? Do I plug in the given x, y values to find C?
have you integrated the other side yet?
I have now: -e^(-y) = e^(sinx) + C
ok so enter in (0,0) and solve for C
I solved that C = 0, So does the equation become -e^(-y) = e^(sinx) ?
-1-1=-2
-1=1+C -1-1=C -2=C
Oh I forgot about the negative on the left.
Then -e^(-y) = e^(sinx) - 2 ?
yes
Thanks!

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