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LECTURE 4. SEPARABLE EQUATIONS
Separable equations. Separable equations are differential equations of the form
dy f (x) (4.1) =. dx g(y)
For example, x + yy�^ = 0 and y�^ = y^2 − 1. A separable equation (4.1) can be written in the differential form as
(4.2) f (x)dx = g(y)dy.
Then, it can be solved formally by integrating both sides of (4.2). We state and prove the rigorous theory of local solutions for (4.2) (and hence (4.1)).
Theorem 4.1. Let f (x) and g(x) be continuous in the rectangle R = {(x, y) : a < x < b, c < y < d}. In addition, if f and g do not vanish simultaneously at any point of R, then (4.2) has one and only one solution through each point (x 0 , y 0 ) ∈ R. The solution is given by x y (4.3) f (x)dx = g(y)dy x 0 y 0 It is essential that f and g do not vanish simultaneously. For example, xdx = −ydy has no solution through the origin.
Proof. Note that f (x) =� 0 for a < x < b or g(y) = 0� for c < y < d. Without loss of generality, we assume g(y) > 0 for c < y < d. Let (^) � (^) x �y F (x) = f (x)dx, G(x) = g(y)dy x 0 y 0 so that (4.3) becomes
(4.4) F (x) = G(x).
Since G�(y) = g(y) > 0 in R, by the inverse function theorem, G−^1 exists and (4.4) can be written
as y = G−^1 (F (x)). That means, dydx^ exists. Then, by differentiating (4.4), we get
F �(x) = G�(x)
dy , or f (x) = g(y)
dy . dx dx This implies (4.2). Moreover, (4.3) gives the initial condition that y = y 0 when x = x 0. To prove the uniqueness, let y be one solution of (4.2) and z be another solution with the same initial condition. Under the hypothesis, the equation
dz
f (x) dx g(z)
implies that dz/dx exists for any (x, z) ∈ R. Let
u = G(y), v = G(z).
Then, du dy dy = G�(y) = g(y) = f (x). dx dx dx
1
dz Similarly, = f (x). Since u and v have the same derivative, they differ by a constant. On the dx other hand, the initial conditions for u and v at x 0 agree. Therefore, u = v everywhere in R. This completes the proof. �
Example 4.2. Consider the initial value problem
(4.5)
dy = 1 + y 2 , y(0) = 1. dx Separating the variables, we write the differential equation as dy = dx. 1 + y^2
Since the constant function never vanishes, upon integration and evaluation, we obtain
tan−^1 y = x + c, tan−^1 1 = c.
Therefore, the (unique) solution of (4.5) is y = tan(x + π/4). The same result is obtained by integrating between corresponding limits y (^) dy x = dx. 1 1 +^ y^20
Orthogonal trajectories. If two families of curves are such that every curve of one family inter sects the curves of the other family at a right angle, then we say that the two families are orthogonal trajectories of each other. For example, the coordinate lines:
x = c 1 , y = c 2
in a Cartesian coordinate system form a set of orthogonal trajectories. Another example is the circles and radial lines r = c 1 , θ = c 2
in a polar coordinate system. Suppose a curve in the (x, y)-plane is such that the tangent at a point (x, y) on it makes an angle φ with the x-axis. The orthogonal trajectory through the same point (x, y) then makes an angle φ + π/ 2 with the x-axis. Since
1 tan(φ + π/2) = − cot φ = − tan φ dy dx dx and since the slope of the curve is dx
= tan φ, we should replace dy
by − dy
in the differential
equation for the original family to get the differential equation for the orthogonal trajectories.
Example 4.3. We consider the family of circles
(4.6) x^2 + y 2 = cx
tangent to the y axis. By differentiating (4.6) and by eliminating c, we obtain a differential equation dy 2 dy x^2 + y 2 = 2x + 2xy dx
, or y − x^2 = 2xy dx that the family of curves (4.6) satisfies. Replace dy/dx by −dx/dy we get the equation of the orthogonal trajectories dx y 2 − x^2 = − 2 xy dy
Lecture 4 2 18.034 Spring 2009
