Separable First Order Differential Equations

Transcription

1 Separable First Order Differential Equations Form of Separable Equations which take the form = gx hy or These are differential equations = gxĥy, where gx is a continuous function of x and hy is a continuously differentiable function of y to guarantee unique solutions. The two forms agree if ĥy = /hy. In differential form we can re-write such an equation as hy = gx thus separating the y dependence from the x dependence. We have alrea encountered the simplest example, i.e., the homogeneous first order linear equation + px y = 0. We can re-write this in the form = px. y Method of Solution Integrating both sides of hy = gx we have y hr dr = x gs ds + c, where c is an arbitrary constant. Then, letting Hy = y hr dr and Gx = x gs ds, we have Hy = Gx + c or Hy Gx c = 0. This is a parametric equation for y in terms of x; if it can be solved for y to give, explicitly, y = yx, c = H Gx + c,

2 then we have an explicit formula for the general solution. Example Consider the differential equation = xy + x + y +. Here we can factor the right hand side: xy + x + y + = x + y +. So we have and, on integration, we have Then, much as before, = x + y + log y + = x + x + y + = ± exp + ĉ. + ĉ x + x + ± eĉ exp = c exp so that the general solution becomes yx, c = c exp x +. = Example This becomes Consider the differential equation = y x + 3 x +. y = x + 3 x +. At this point we need to pause and provide a

3 Reminder: Partial Fractions Decomposition When we need to integrate a function which is a quotient of two polynomials, i.e., fx = b x n + b x n + + b n a 0 x n + a x n + a n x + a n, a 0 0, we first need to transform fx into another form. This is done by the procedure of partial fractions decomposition. The first step is to factor the denominator. Assuming it has distinct roots r, r, r n the denominator takes the form a 0 x r x r x r n. We then try to write c fx = + c + + c n. x r x r x r n Recombining the right hand side into a single fraction with a common denominator we have fx = c x r x r n + + c n x r x r n. x r x r x r n The numerator shown here, and the expression a 0 b x n + b x n + + b n must be the same. Equating the coefficients of corresponding powers of x gives n linear equations in the n unknowns c, c., c n which can be solved to obtain the desired expression c x r + c x r + + c n x r n. If one of the roots, say r, is a double root, we try instead for an expression fx = c x + d x r + c x r + + for a triple root we use c x + d x + e x r 3, etc. c n x r n ; As an example we construct the partial fractions decomposition of fx = x x + x x + 8 x + 4 = x x + x + x +. 3

4 Since r = is a double root, we try to achieve the form fx = c x + d x + + c x +. Recombining these two fractions into a single fraction we obtain fx = c x + d x + + c x + x + x + = c + c x + c + d + 4c x + d + 4c x + x +. Comparing this with the original formula for fx and equating coefficients of corresponding powers of x we arrive at three equations: i : c + c = ; ii : c + d + 4 c = ; iii : d + 4 c =. Substituting equation iii into equation ii we find c = 3. Using this in equation i we obtain c = 4 and then using this in equation iii we find d = 4. Accordingly, we have fx = x x + x x + 8 x + 4 = 3 x x + x +. Example Continued We return to our unfinished differential equations example. To apply partial fractions decomposition to that case we note that x + 3 x + = x + x +. So we try x + 3 x + = c x + + c x + = c x + + c x + x + 3 x + = c + c x + c + c. x + 3 x + 4

5 Thus we need c + c = 0, c + c =. This is easily solved to give c =, c =. Consequently we may now rewrite our differential equation in the form Integrating, we have y = x +. x + y = log x + log x + + ĉ. Renaming ĉ as log c, we have x + y = log. c x + Again absorbing the signs into the constant c we have, renaming c if necessary, yx, c =. log x + c x + We cannot always assume that we will be able to solve the integrated equation Hy = Gx + c to get an explicit formula for the general solution yx, c. Example 3 We consider the differential equation = sinx logy. Rewriting this as logy = sinx and integrating we obtain y logy y = cosx + c, y > 0 which is not directly solvable in the form y = yx, c. This, of course, is one of the reasons why numerical approximation methods are as 5

6 valuable as they are. The equation shown is far from useless, however. For example, suppose we want the value of c corresponding to the initial condition y0 =. Substituting y = and x = 0 into the implicit formula shown we have log = cos0 + c c = log. The solution yx of the initial value problem is then the solution of the equation hx, y = y logy y + cosx log + = 0, an example of what we have called a parametric equation, and we can obtain individual values of yx by substituting values of x into this equation and then solving numerically, by Newton s formula or the fixed point iteration method, for y. Thus if we take x =., start with the value obtained from the initial condition at y = 0 and apply Newton s method y k+ = y k y k logy k y k + cos. log + / logy k we obtain almost instantly y. =.007. Then with x =. and starting value.007 we obtain y. =.085. We can continue in this way to build a table of values for the solution near x =. Homogeneous First Order Equations to be homogeneous of degree n if A function fx, y is said fαx, αy α n fx, y for all values of x, y and α. A first order differential equation = fx, y is said to be of homogeneous type if fx, y is homogeneous of degree 0, i.e., fαx, αy α 0 fx, y = fx, y. for all values of x, y and α. 6

7 Example 4 then fαx, αy If fx, y = xy + y xy = αxαy + αy αx αy = α xy + y α x y = xy + y x y = fx, y, showing fx, y to be homogeneous of degree zero in this case. The present homogeneous case should not be confused with the linear homogeneous case where fx, y = px y; indeed, if it were true that pαx αy = px y, then we would have pαx = α px and since this must be true for all x and α, we would necessarily have, with α replaced by x and x replaced by : px = px = x p which implies px = p x, i.e., px is a multiple of. This would x correspond to the case = a x y for which we alrea know that the solution is yx, c = c expa log x = x a. Method of Solution Equations of homogeneous type can be solved, at least in principle, by making use of a change of variable which reduces the equation to a separable first order equation. 7

8 Starting with fx, y homogeneous of degree zero in the differential equation = fx, y we set y = x v and obtain Then or d x v = x dv x dv hv dv = + v = fx, x v = f, v. = f, v v, f, v v dv = x. Integrating, we have, with Hv an antiderivative of hv = Hv = log x + ĉ = log x + log c = logc x. f,v v, Assuming we can find an inverse function, H, for Hv, we obtain v = H logc x and then, since y = x v, we have as the general solution yx, c = x H logc x. Example 5 Setting y = x v, we have Consider the differential equation x dv + 4x + 3y 3x + y = 0. + v = 4 + 3v 3 + v. 8

9 Transposing v and adding fractions we have x dv = 4 + 3v 3 + v v = 4 + 3v + v3 + v 3 + v = v + 6v + 4 x dv v + v + =. 3 + v 3 + v Changing to differential form this is 3 + v v + v + dv = x. Then, applying the method of partial fractions, Integrating, we have This gives v + + v + dv = x. log v + + log v + = log x + ĉ. log v + v + = log cx = log c x. Thus we have v + v + = ± c x. Letting a = ± c be a constant of arbitrary sign, we have v + 3 v + + a x = 0. Solving this quadratic equation for v in terms of x and a, vx, a = 3 ± a x 9

10 Remembering that y = x v, in terms of the original dependent variable y the solution is yx, a = x 3 ± 4a However, it would be misleading to give the impression that matters always work out so that the solution can be obtained in closed form. In fact this is often not the case as we see in the next example. Example 6 Setting y = x v we have or x dv = + v v This is the same as We consider the very simple differential equation = x + y x y. x. x dv + v = x + x v x x v = + v v v = + v v v v v + v dv = x or + v dv v + v dv = x. Then, integrating, we have tan v log + v = log x + c. = + v v. In this case there is essentially no hope of solving for v to obtained v = vx, c with vx, c an expression in terms of elementary functions; the best we can do is to say that the general solution yx, c = x vx, c satisfies the parametric equation tan y x log + y 0 x = log x + c.

11 QuickCheck Exercises. Find the solution of the initial value problem = + y x π, y 3π 4. Find the general solution of = x + y xy + x =. as the solution of a parametric equation hx, y, c = 0. Then find the value of c corresponding to the initial condition y =. Find a four decimal place approximation to y.. 3. Find the solution of = y y x satisfying the initial condition y = The velocity, vt, of a certain mass sliding along a surface with friction satisfies dv dt + v + v = 0. If the initial velocity is v0 = 0 meters per second, when is the velocity reduced to meter per second?