Separable Differential Equations

1. Nov 3, 2008

Moridin

1. The problem statement, all variables and given/known data

I'm having trouble understanding my class notes from a lecture on separable differential equations.

I would like to solve the equation g(y)y' = f(x)

3. The attempt at a solution

g(y)y' = f(x), G(x), F(x) exists and are continuous

The left side is the derivative of G(y(x)) and the right is F(x) + C

$$\frac{d}{dx} G(y(x)) = \frac{d}{dx} F(x) + C$$

G'(y(x))y'(x) = g(y(x))y'(x)

$$\frac{d}{dx} (F(x) + C) = F'(x) + 0 = f(x)$$

G(y(x)) = F(x) + C

So do you simply do

G-1(G(y(x))) = y(x) = G-1(F(x)) + G-1(C) ?

2. Nov 3, 2008

gabbagabbahey

Assuming $G^{-1}$ exists and is a 1:1 map, then yes....Some functions do not have inverses that uniquely determine y(x), for instance $y^2(x)=F(x)+C \Rightarrow y(x)= \pm \sqrt{F(x)+C}$ which is not a single valued function.

3. Nov 3, 2008

bobmerhebi

hello, I'm also taking a class on ODE but i have a problem -i use An Intro course in Diff. eq.'s by Zill - that i get a nonsense result here is the eq:

sin3x + 2y(cos3x)^3 = 0 (here ^ is to raise a power.how are u raising powers & all the mathematical writings?)

the eq in standard form look: (y^2)'=2ydy= -sin3x dx/2(cos 3x)^3.

the last result i get which is nonsense ofcourse is: y^2 = -1/6(cos3x)^2. another result includes tan3x but is still negative.

so y^2 is negative which is impossible. is the result right? I think there's a problem with the D.E. given.

hope u can help. thx