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

1. Jan 18, 2006

### mr_coffee

Hello everyone, I am confused on what they are wanting me to do here on this problem:

has an implicit general solution of the form

In fact, because the differential equation is separable, we can define the solution curve implicitly by a function in the form

Find such a solution and then give the related functions requested.

here is my work, i think they wanted me to do, they gave no intial condition so i solved for C:

but it was incorrect

2. Jan 18, 2006

### 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.

Last edited: Jan 18, 2006
3. Jan 18, 2006

### 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?

4. Jan 18, 2006

### HallsofIvy

Staff Emeritus
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 5e2x- arcsin(x/2)= K. Can't you just look at that and pick G and H?

5. Jan 18, 2006

### benorin

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

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

6. Jan 18, 2006

### mr_coffee

Ahhh, thanks so much everyone it was correct. weee