Recall what is an 1st-order ODE Implicit form F(x, y, y’) = 0 For example: x3y’– 4y2 = 0 Explicit form y’ = f(x, y) For example: y’ = 4x3y2 Here y = y(x) and y’ = dy/dx We already learned solving it by separation of variable. Now we will learn a different technique by utilizing the concept of partial differentiation (偏微分).

From this it follows that if u(x, y) = c = const, then du = 0. In calculus we know that if a function u(x, y) has continuous partial derivatives, its differential (also called a total derivative) is From this it follows that if u(x, y) = c = const, then du = 0. For example, if u = x + x2y3 = c, then du = (1+2xy3)dx + 3x2y2dy = 0 or This is an ODE we can solve by going backward. 對x偏微分 對y偏微分

[Continue from previous slide.] Suppose we have the following ODE: Observe that the above ODE can be written as (1+2xy3)dx + 3x2y2dy = 0 Or M(x, y) dx + N(x, y) dy = 0 (1)

Only if we can find u(x, y). A first-order ODE written as (1) M(x, y) dx + N(x, y) dy = 0 is called an exact differential equation (正合微分方程) if the differential form M(x, y) dx + N(x, y) dy is exact. That is, this form is the differential (2) of some function u(x, y). Then (1) can be written du = 0. By integration we immediately obtain the general solution of (1) in the form (3) u(x, y) = c. Only if we can find u(x, y).

Recall that (4) Therefore (5) For an equation of the form M(x, y) dx + N(x, y) dy = 0, this condition indicates that the differential equation is exact. Once the differential equation is exact, we will be able to write the solution to be u(x, y) = c. (The problem now is how to find this function u(x, y) = c.) Or

M(x, y) dx + N(x, y) dy = 0 (1) (4) where M(x,y) and N(x,y) are known. If (1) is exact (Eq (5) from previous slide), the function u(x, y) can be found in the following systematic way. From (4a) we have by integration with respect to x (6) u = ∫M dx + k(y) In this integration, y is regarded as a constant, and k(y) plays the role of a “constant” of integration. To determine k(y), we derive u/y from (6), use (4b) to get dk/dy, and integrate dk/dy to get k(y).

in this integration, l(x) is to be regarded as a constant. (4) where M(x,y) and N(x,y) are known. One can alternatively begin from 4(b) with integration with respect to y (6*) u = ∫N dy + l(x) in this integration, l(x) is to be regarded as a constant. To determine l(x), we derive u/x from (6*), use (4a) to get dl/dx, and integrate dl/dx to get l(x). Function u(x, y) determined by either approach, of course, will be the same.