Diff.Eq. Seperation of variables.

Jtechguy21
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Homework Statement


Solve the given differential equation subject to the indicated initial condition.

(e^-y + 1)sinxdx=(1+cosx)d, y(0)=0

Homework Equations


Basically we have to use separation of varaibles to solve before using initial value condition.

The Attempt at a Solution


After separation of variables

Dy/(e^-y +1) = sinx dx/(1+cosx)

take the integral of both sides
∫Dy/(e^-y +1)=ln|e^-y+1|+y

∫sinx dx/(1+cosx)= -ln(1+cosx)+c

Clean it up a bit
ln|e^-y+1|+y= -ln(1+cosx)+cI have no idea what to do now with the y(0)=0
That means plus in x=0 for the equation correct?
 
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You use the initial condition to determine the value of c.

when x = 0, y = 0.
 
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SteamKing said:
You use the initial condition to determine the value of c.

when x = 0, y = 0.

How did you get y=0?
 
Jtechguy21 said:
How did you get y=0?

That's what y(0) = 0 means!
 
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well still there is a problem if x(0) is not zero for y(0)=0...
In that case you solve c with respect to x(0) (a parameter) and input it
 
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okay thank you.
My last question is.
can someone please check my integration
∫Dy/(e^-y +1)=ln|e^-y+1|+y

to make sure ^ is correct?
 
you can always try to take the derivative of the righthand side and check by yourself ;) much easier
 
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