Elementary ODE Review First Order ODEs First Order Equations Ordinary differential equations of the fm y F(x, y) () are called first der dinary differential equations. There are a variety of techniques to solve these type of equations and main methods are: (i) (ii) (iii) (iv) (v) (vi) (vii) (viii) separable linear Bernoulli Ricatti homogeneous linear fractional exact Legendre transfmations. Separable Equations A separable first der differential equation has the fm: f (x)g(y) (2) The general solution is found by separating the differential equation and integrating including a single constant of integration, i.e. g(y) f (x) + c. F example, to solve y xy + x + 2y + 2,
it is necessary to rewrite it as y (x + 2)(y + ). Separating variables and writing in integral fm gives y + x + 2, and integrating yields ln y + x2 2 Letting c ln(k) and solving f y gives + 2x + c. y k e x2 /2+2x. As a second example, consider Separating variables gives and integrating yields Letting c 2 ln(k) and solving f x gives Imposing the initial condition gives which gives on solving f k which gives the solution dt x(2 x), x(0) x 0. x(2 x) dt, (ln x ln 2 x ) t + c. 2 x x 0 2ke2t + ke 2t. 2k + k, k x 0 2 x 0, x 0 2, x(t) 2x 0 e 2t 2 x 0 + x 0 e 2t. In the case where x 0 2, then the solution is x(t) 2 f all t. 2
.2 Linear Equations Equations of this type are in the fm To solve this, we introduce the integrating fact + p(x)y q(x). (3) µ e p(x). (4) This is created so that when both sides of (3) are multiplied by µ, the left side (3) is a derivative of a product, that is, it becomes ( ) µ + py d (µy), and then (3) can be integrated. F example, if then p(x) 2 x, so that µ On multiplying (5) by µ gives 2y x 2x3, (5) e 2 x e 2 ln x x 2. x 2 2y x 3 2x x 2, which simplifies to Integrating gives and solving f y gives ( ) d x 2 y 2x x 2. x 2 y x2 + x + c, y x 4 + x + cx 2..3 Bernoulli Equations of the fm + p(x)y q(x)yn (n 0, ) (6) 3
are called Bernoulli equations. Dividing both sides of (6) by y n gives Let v, then dv y n into (7) gives y n + p(x) q(x) (7) yn ( n) y n dv ( n) y n dv + p(x)v q(x) n. Upon making this substitution which is linear. So Bernoulli equations can be reduced to linear equations. Example Consider y 2x y3. This is an example of a Bernoulli equation where n 3. Putting this into standard fm gives Letting v y 2, then dv 2 y 3 y 3 2x y 2 (8), and (8) is transfmed to 2 dv 2x v, dv + v x 2. As this is linear, then p(x) x, and the integrating fact f this is x, so that x dv + v d (xv) 2x, and thus So that xv x 2 + c, v x + c x. y 2 x + c x, y ± x + c x. 4
.4 Ricatti Equations Ricatti equations have the fm: a(x)y2 + b(x)y + c(x). (9) To find a general solution to this requires having one solution first. Given this solution, it is possible to change equation (9) to a linear equation. If we let y y 0 + u, where y 0 is a solution to (9), then (9) is transfmed to the linear equation u (2a(x)y 0 + b(x)) u a(x), which is linear. To illustrate, we consider the following example y2 x 2 + y +. (0) x Since y 0 x is a solution to this equation, let y x + u, and (0) becomes u u 2 (x 2 x 2 + 2 xu + u ) 2 + ( x + ) +. x u Simplifying gives u u 2 xu x 2 u 2, and multiplying by u 2 and rearranging gives rise to the linear equation u x u x 2. () Here p(x) x so this has the integrating fact µ e x and upon integration gives Since y x + u, this gives y as y x + u x u x 2 d ( u ) x x 3, u x 2x 2 + c, u cx 2x. cx 2x. x, so () becomes 5
.5 Homogeneous Equations Equations of the fm ( y ) F x are called homogeneous equations. Substituting y xu will yield the equation (2) which separates to x du + u F (u). du F(u) u x. Consider the previous example, If we let y xu, then (3) becomes y2 x 2 + y +. (3) x d(xu) u 2 + u +, and simplifying and separating gives Integrating gives Since y xu this gives solving f y leads to the solution x du + u u2 + u + du u 2 x. 2 ln u + u ln x + ln c, 2 u + u cx2. y x + y x cx2, y + x y x cx2, y cx3 + x cx 2. 6
.6 Linear Fractional Equations that have the fm are called linear fractional. Under a change of variables, ax + by + e cx + + f, (4) x x + α, y ȳ + β, we can change equation (4) to one that is either homogeneous (if ad bc 0) to one that is separable (if ad bc 0). The following examples illustrate. Consider If we let 2x 3y + 8 3x 2y + 7. (5) x x + α, y ȳ + β, then (5) becomes Choosing dȳ d x 2 x 3ȳ + 2α 3β + 8 3 x 2ȳ + 3α 2β + 7. 2α 3β + 8 0, 3α 2β + 7 0, (6) leads to dȳ d x 2 x 3ȳ 3 x 2ȳ, (7) a homogeneous equation. The natural question is, does (6) have a solution? In this case, it does and we can find that the solution is α and β 2. The solution of (7) is x 2 3 xȳ + ȳ 2 c (8) and from our change of variables (x x, y ȳ + 2) we find the solution to (5) is As a second example, consider (x + ) 2 3(x + )(y 2) + (y 2) 2 c. (9) 4x 2y + 8 2x y + 7. (20) If we let x x + α, y ȳ + β, 7
then We choose dȳ d x 4 x 2ȳ + 4α 2β + 8 2 x ȳ + 2α β + 7. 4α 2β + 8 0, 2α β + 7 0, but this has no solution! However, if we let u 2x y, then (20) becomes which is separable! Its solution is given by du 6 u + 7, u 2 + 4u 2x + c, and upon back substitution, we obtain the solution of (20) as (2x y) 2 + 4(2x y) 2x + c,.7 Exact Equations An dinary differential equation of the fm F(x, y), (2) has the alternate fm M(x, y) + N(x, y) 0. (22) If M and N have continuous partial derivatives of first der in some region R and M y N x, then the ODE (22) is said to be exact and can be integrated by setting φ x M, and φ y N. F example, consider the differential equation 2xy x 2 + y 2, (23) which can be written as 2xy + (x 2 + y 2 ) 0. (24) 8
If we identify that M and N are so so (24) is exact. Setting and integrating the first gives M 2xy and N x 2 + y 2, φ x M y 2x N x, 2xy, and φ y x2 + y 2, φ x 2 y + g(y), taking the partial of this with respect to y gives Comparing this to φ y x2 + y 2 gives that φ y x2 + g (y). g (y) y 2, so so that From Cal III we know that but in this case this is g(y) y3 3 + c, φ x 2 y + y3 3 + c. dφ φ x + φ y, dφ 2xy + (x 2 + y 2 ) 0 so φ is a constant. Thus we have as solutions to 2xy + (x 2 + y 2 ) 0 φ k x 2 y + y3 3 + c k, and absbing the constant k into c gives as the set of possible solutions to (23). x 2 y + y3 3 + c 0 9
.7. Legendre Transfmations Sometimes it is necessary to solve me general equations of the fm F(x, y, y ) 0, (25) say, f example y 2 xy + 3y 0. (26) One possibility is to introduce a contact transfmation that enables one to solve a given equation. Contact transfmations, in general, are of the fm with the contact condition that x F(X, Y, Y ), y G(X, Y, Y ), y H(X, Y, Y ), (27) G X + G Y y + G Y Y F X + F Y y + F Y Y H. One such contact transfmation is called a Legendre transfmation and is given by One can verify that x dy dx, Substitution of (28) in (26) gives a linear ODE! Solving gives ( d dx Substituting (30) back into (28) gives Solving the first of (3) f X 2 gives d dx y X dy dx Y, y X. (28) X dy ) dx ( ) Y dy dx X d2 Y dx 2 d 2 Y dx 2 X. 2X dy dx 3Y + X2 0 (29) Y CX 3 2 X 2. (30) x 3 2 cx 2 2X, y 2 cx 3 2 X 2. (3) X 3c ± 9c 2 + 32x 2, (32) 8 and from the second of (3) gives ( y c 3c ± ) 3 ( 9c 2 + 32x 3c ± ) 4 9c 2 + 32x, 2 8 8 the exact solution of (26). 0