MIT OpenCourseWarehttp://ocw.mit.edu 18.034 Honors Differential Equations Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.� � 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� = y2 − 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 (x0, y0) ∈ R. The solution is given by x y (4.3) f(x)dx = g(y)dy x0 y0 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 x0 y0 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, dy exists. Then, by differentiating (4.4), we get dx 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 = y0 when x = x0. 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� � dzSimilarly, = 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 x0 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 + y2 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 limitsy dy x = dx.1 + y2 1 0 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 = c1, y = c2 in a Cartesian coordinate system form a set of orthogonal trajectories. Another example is the circles and radial lines r = c1, θ = c2 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 1tan(φ + π/2) = − cot φ = −tan φ dy dx dxand 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 + 2xydx, or y − x 2 = 2xydx 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 = −2xydy. Lecture 4 2 18.034 Spring 2009We write it in differential form as 2xydx − x 2dy + y 2dy = 0. Multiplying by 1/y2 then gives∗ � �2xd + dy = 0, y 2xand hence, + y = c. We arrange it into y x 2 + y 2 = cy, which represents a family of circles tangent to the x-axis. Although the analytical steps require x = 0� , y = 0� , and y2 =� x2, the final result is valid without these restrictions. The quantity c defined by x2 + y2 = cx is constant on the original curves of the family, but not on the orthogonal trajectories. That is why c must be eliminated in the first step. Exercise. Show that the orthogonal trajectories of the family of geometrically similar, coaxial el-lipses x 2 + my 2 = c, m > 0 mare given by y = ±|x| . Exercise. Show that the solution curves of any separable equation y� = f(x)g(y) have as orthogo-nal trajectories the solution curves of the separable equation y� = −1/f(x)g(y). ∗This procedure makes the equation exact and the solution is defined implicitly. The factor 1/y2 is called a integrating factor. We will study exact differential equation more systematically later. Lecture 4 3 18.034 Spring
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