Section 2.3
Separable Equations

Given a general first order D.E.

$$\frac{dy}{dx} = f(x,y)$$

we rewrite it in the form

M(x,y)dx + N(x,y)dy = 0

Now, if M is purely a function of x, and N is purely a function of y, then we may simply write:

M(x)dx + N(y)dy = 0

Let

H₁'(x) = M(x), H₂'(y) = N(y)

now we can rewrite our D.E. again as

$$H_1'(x) + H_2'(y)\frac{dy}{dx} = 0$$

by the chain rule we can obtain

$$\frac{d}{dx}H_2(y) = H_2'(y)\frac{dy}{dx}$$

and so we get (noting that $H_1'(x) = \frac{d}{dx}H_1(x)$)

$$\frac{d}{dx}[H_1(x) + H_2(y)]$$

and now we integrate (Note: with respect to x) to obtain

H₁(x) + H₂(y) = c.

Question: what would have happened if we just integrated the equation

M(x)dx + N(y)dy = 0?

Answer: nothing terrible

$$\int{M(x) dx} + \int{N(y) dy} = \int{0 \: d?}$$