Differential equation, confused on directions, F(x,y)?

[mr_coffee]

Hello everyone, I am confused on what they are wanting me to do here on this problem:
http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/8e/03f256344c773f4cb31a6c00deae771.png
has an implicit general solution of the form
http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/46/676cc60ad18cc49904f5f4ac4fa61e1.png
In fact, because the differential equation is separable, we can define the solution curve implicitly by a function in the form
http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/82/62c381e72069d44ef291f5391ed7871.png
Find such a solution and then give the related functions requested.
http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/57/f15a6cea760ce4906a6ec16f9d29451.png
here is my work, i think they wanted me to do, they gave no intial condition so i solved for C:
http://img39.imageshack.us/img39/1227/lastscan5vx.jpg

but it was incorrect

 

[Benny]

Well what is supposed to be the correct answer? I can't see much wrong with what you've done.

Edit: You seem to have replaced the y with x after the integration of the exponential.

 

[mr_coffee]

thats a good question, it doesn't tell me. it will tell me if i get it right, not if i get it wrong. Whats this implicit general solution stuff do i have to take partial derivatives or somthing?

 

[HallsofIvy]

WHY do you keep writing $\frac{1}{e^{-2x}}$? Surely you know that $\frac{1}{e^{-2x}}= e^{2x}$

Your instructions were to write the answer as F(x,y)= K and then write it as F(x,y)= G(x,y)+ H(x,y)= K

You already have 5e^2x- arcsin(x/2)= K. Can't you just look at that and pick G and H?

 

[benorin]

Going from the second line to the third line your "y" became an "x". As far as what they want:

Hello everyone, I am confused on what they are wanting me to do here on this problem:
http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/8e/03f256344c773f4cb31a6c00deae771.png
has an implicit general solution of the form
http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/46/676cc60ad18cc49904f5f4ac4fa61e1.png
In fact, because the differential equation is separable, we can define the solution curve implicitly by a function in the form
http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/82/62c381e72069d44ef291f5391ed7871.png
Find such a solution and then give the related functions requested.

They asked for $$F(x,y)=G(x)+H(y)=K$$

So "Find such a solution" is satisfied by $$F(x,y)=-\mbox{arcsin}\left( \frac{x}{2}\right) + 5e^{2y} = K$$

but the answer they want is "give the related functions requested"

so try $$G(x)= -\mbox{arcsin}\left( \frac{x}{2}\right) \mbox{ and }H(y)= 5e^{2y}$$

 

[mr_coffee]

Ahhh, thanks so much everyone it was correct. weee