Further Ordinary Differential Equations

Transcription

1 Advanced Higher Notes (Unit ) Prerequisites: Standard integrals; integration by substitution; integration by parts; finding a constant of integration; solving quadratic equations (with possibly complex roots); solving simultaneous equations. Maths Applications: Solving differential equations. Real-World Applications: Electrical circuits; damped/forced motion. Ordinary Differential Equations Special types of differential equations (DEs) have been introduced earlier in this course and in Higher. Here, we shall look at more general classes of differential equations. Definition: An n th order ordinary differential equation (ODE) is an equation containing (i) an unknown function of a single variable and (ii) derivatives of the function up to the n th order. Most DEs are difficult to solve exactly. Fortunately, many DEs that arise in real-life fall into categories that we can easily stu. One of these, of main interest to us, is the following type. Definition: An n th order linear ODE is one of the form, n = (r) a ( x ) = f (x) r ( r ) r 0 For example, x 11 cot x + 98 y = π sin x M Patel (April 01) 1 St. Machar Academy

2 Advanced Higher Notes (Unit ) In this course, the focus will be mainly on 1 st and nd order linear ODEs. We have encountered nonlinear DEs. Most separable DEs are nonlinear. Definition: An n th order linear ODE with constant coefficients is of the form, with all the a constant. r n = (r) a = f (x) r ( r ) r 0 Definition: A linear ODE is homogeneous if the function f (x) in the above definition is identically 0. A linear ODE is non-homogeneous (aka inhomogeneous) if it is not homogeneous. Definition: The general solution of a homogeneous differential equation is called the Complementary Function (CF). The term ordinary refers to the fact that the unknown function depends on only 1 (independent) variable; many interesting DEs involve functions of several variables, in which case we can form a partial differential equation (PDE). Also, there is often more than 1 DE describing a particular phenomenon, in which case we have a system of differential equations. Examples from physics include: Maxwell s equations of electromagnetism, which form the basis for all electrical and magnetic technology, are a system of 8 linear nd order PDEs in 4 independent variables and unknown functions; Einstein s field equations of gravitation, which describe gravity as a warping of space and time around any object, are a system of 10 nonlinear nd order PDEs in 4 independent variables and 10 unknown functions; Schrödinger s wave equation, which forms the modern M Patel (April 01) St. Machar Academy

3 Advanced Higher Notes (Unit ) basis for atomic physics (including phenomena such as teleporting photons), is a linear nd order PDE in or 4 independent variables and 1 (complex!) unknown function; the Dirac equation, a relativistic version of the Schrödinger equation which predicts the existence of antimatter, is a linear 1 st order PDE in 4 independent variables and 1 (complex) unknown function. An example from biology are the Lotka-Volterra predator-prey equations, which describe the interaction of a predator and a prey, are a system of nonlinear 1 st order ODEs in 1 independent variable and unknown functions. Note that many real-life differential equations are of 1 st or nd order. A notable example of a rd order nonlinear PDE in independent variables and 1 unknown function is the Kortewig-de Vries equation, which describes shallow water waves (solutions of this equation are called solitons - a wave that keeps its shape while travelling at constant speed - and were seen in the Union Canal in Scotland by John Scott Russell in the 180s during his investigations into efficient canal boat designs). 1 st Order Equations and the Integrating Factor Previously, 1 st order homogeneous equations were studied. We will now solve the most general type of 1 st order linear ODE. Theorem: A 1 st order linear ODE can be written in the form, + P (x) y = f (x) and is solved by multiplying both sides of this equation by the Integrating Factor (IF) e P ( x ) and then integrating both sides with respect to the independent variable (for this DE, that variable is x). Before multiplying by the IF, it is crucial to rewrite the DE in the above form (if it s not alrea in that form). The term must be written on its own. M Patel (April 01) St. Machar Academy

4 Advanced Higher Notes (Unit ) Example 1 y = x x The function P (x) equals x. The IF is, e x = e 1 x = e ln x + D = ln + D = ln e D = e D x where D is the constant of integration. Multiplying the above DE by the IF gives, e D x e D x y = x. e D x x e D x e D x 6 y = e D x 4 As the constant e D is never 0 (as it s always positive), we can divide by it to get, x 6 x y = x 4 This canceling of the constant always occurs in every type of IF DE; if it doesn t happen, a mistake has been made somewhere. Once this step has been reached correctly, the rest is routine. First, rewrite the LHS using the product rule as, d ( x y ) = x 4 M Patel (April 01) 4 St. Machar Academy

5 Advanced Higher Notes (Unit ) Finally, integrate both sides with respect to x (and, for the LHS, remembering that integration is the opposite of differentiation) and solve for the dependent variable, x y = x + C y = x + C x Time-permitting, it should be checked that this alleged solution does satisfy the DE. Example cos x + y sin x = 1 Rewriting this equation gives (assuming cos x 0 ), + y tan x = sec x The IF is, e tan x = e ln cos x + D = D e cos x Then, D e cos x + D e cos x y tan x = D e cos x sec x sec x + sec x tan x y = sec x Hence, Thus, d ( y sec x ) = sec x M Patel (April 01) St. Machar Academy

6 Advanced Higher Notes (Unit ) y sec x = tan x + C y = sin x + C cos x Example + y = The IF is, e = e D So, + e x y = e x d ( e x y ) = e x y = + C y = C + The worst-case scenario with DEs involving an Integrating Factor occurs when integration by parts must be used on the RHS. The following example is a slight modification of Example. Example 4 + y = x The IF is the same as before, e = e D M Patel (April 01) 6 St. Machar Academy

7 Advanced Higher Notes (Unit ) So, + e x y = x d ( e x y ) = x y = x Integrating the RHS of this equation by parts gives (do it!), y = x 49 + C y = x 49 + C Second-Order Equations The second order ODEs that we will stu will be homogeneous and nonhomogeneous, but only those with constant coefficients. Homogeneous Remember that the general solution of a homogeneous differential equation is given by the Complementary Function (CF). There are possible forms for the CF for such a DE, depending on the nature and number of the roots of a quadratic equation formed directly from the DE. Theorem: The general solution of a homogeneous nd order linear ODE with constant coefficients, a + b + c y = 0 is obtained by solving the associated Auxiliary Equation (AE), M Patel (April 01) St. Machar Academy

8 Advanced Higher Notes (Unit ) a p + b p + c = 0 (p C ) If the auxiliary equation has real (distinct) roots p and p, then the CF 1 is of the form, 1 y = A e p x + B e p x (A, B R ) CF If the auxiliary equation has 1 real (repeated) root p, then the CF is of the form, y = (A + B x) e px (A, B R ) CF If the auxiliary equation has a pair of complex (conjugate) roots p = r 1 + is and p = r is, then the CF is of the form, y = e rx ( A cos sx + B sin sx ) (A, B R ) CF Example Find the general solution of y = 0. The auxiliary equation is, p + 8 p + 1 = 0 Solving this gives distinct, real solutions p = and p = 6. The 1 CF, and thus the general solution (since we have a homogeneous equation), is thus, y = A CF e x + B 6 e x Example 6 Find the general solution of 10 + y = 0. The auxiliary equation is, M Patel (April 01) 8 St. Machar Academy

9 Advanced Higher Notes (Unit ) p 10 p + = 0 Solving this gives 1 real repeated solution p =. The general solution is thus, y = (A + B x) CF Example Find the particular solution of y = 1 and = 0 when x = y = 0 satisfying The DE must be rewritten as, y = 0 The auxiliary equation is, p 6 p + 14 = 0 The solutions are p = + i and p = i. The general solution 1 is thus, y = CF ( A cos x + B sin x ) To find the particular solution, we differentiate the CF, CF = ( A cos x + B sin x ) x + e ( A sin x + B cos x ) Substituting x = 0, y = 1 into the first equation and x = 0, = 0 into the seconield respectively, M Patel (April 01) 9 St. Machar Academy

10 Advanced Higher Notes (Unit ) ( A. B. ) = 1 ( A. B. ) ( A. B. ) = 0 These first of these gives A = 1 and the second then gives B =. Hence, the required particular solution is, y = PS Non-Homogeneous x e cos x sin x Definition: A Particular Integral () is a solution of the inhomogeneous DE. Theorem: The general solution of an inhomogeneous nd constant coefficients, order linear ODE with a + b + c y = f (x) is obtained by adding the Complementary Function to a Particular Integral, y = G y + y CF The is chosen to be of the same form as f (x). The following table summarises the situation. f (x) C x + D C x + D x + E C e px C sin px C cos px S x + T S x + T x + U S e px S sin px + T cos px S sin px + T cos px M Patel (April 01) 10 St. Machar Academy

11 Advanced Higher Notes (Unit ) Slight variations of the last s are important too. f (x) C e px + D S e px + T C sin px + D S sin px + T cos px + U C cos px + D S sin px + T cos px + U All constants in the LHS of the above tables will be known from the form of f (x). The unknowns in the RHS of the above tables are S, T, and U. The proposed is substituted into the inhomogeneous equation to find the unknown constant(s); thus, the will then have no arbitrary constants. Example 8 Find the general solution of y = x 4. The CF for the associated homogeneous DE was obtained in Example as, y = A CF e x + B 6 e x For the try, y = S x + T Then, y = S y = 0 Substituting these pieces of information into the inhomogeneous DE gives, 0 + 8S + 1 (S x + T ) = x 4 (1S ) x + (8S + 1T ) = x 4 M Patel (April 01) 11 St. Machar Academy

12 Advanced Higher Notes (Unit ) Equating coefficients of x and solving gives S = 1 4 and T = 1. Substituting these values into the gives, y = 1 4 x 1 Finally, the general solution is, y = A G e x + B x 1 Example 9 Find the general solution of 10 + y = x. Solving the associated homogeneous DE, 10 + y = 0 from Example 6 gives, y = (A + B x) CF For the try, y = S x + T x + U Note that even though the RHS of the non-homogeneous DE has no x term, it must still be included in the proposed (differentiating th term in the will give a constant, which contributes to the constant on the RHS). Then, y = S x + T y = S M Patel (April 01) 1 St. Machar Academy

13 Advanced Higher Notes (Unit ) Substituting these ingredients into the inhomogeneous DE gives, S 10 (S x + T ) + (S x + T x + U ) = S x + (T 0S ) x + (S 10T + U ) = x x Equating coefficients gives, S = 1 T 0S = 0 S 10T + U = Solving these equations gives S = 1, T = 4 1 and U = Hence, y = 1 x x 44 1 The general solution is, y = (A + B x) G + 1 x x 44 1 Example 10 Find the general solution of y = 4 sin x. Solving the associated homogeneous from Example gives, y = e CF ( cos + sin ) x A x B x For the try, y = S sin x + T cos x M Patel (April 01) 1 St. Machar Academy

14 Advanced Higher Notes (Unit ) Then, y = S cos x T sin x y = S sin x T cos x Substituting these into the inhomogeneous DE gives, ( S sin x T cos x) 6 (S cos x T sin x) + 14 (S sin x + T cos x) = 4 sin x Collecting the sine and cosine terms together gives, after a bit of work, ( 11S + 0T ) sin x + ( 0S 11T ) cos x = 4 sin x Equating the sine and cosine terms gives the simultaneous equations, 11S + 0T = 4 0S 11T = 0 whose solution is S = and T = Hence, The general solution is, y = sin x cos x y = e G ( cos + sin ) x A x B x sin x cos x M Patel (April 01) 14 St. Machar Academy

15 Advanced Higher Notes (Unit ) Example 11 Find the particular solution of 4 satisfying the initial conditions y = 1 and The CF is, + 4 y = + 8 = 0 when x = 0. y = (A + B x) CF For the try, y = S + T Then, y = S y = 9S Substituting these into the inhomogeneous DE and simplifying gives, (9S 1S + 4S ) + 4T = S + 4T = Thus, S = 1 and T =. So, the is, y = + The general solution is thus, y = (A + B x) G + + To find the particular solution, we must differentiate the general solution, y = (A + B ) G + B x + M Patel (April 01) 1 St. Machar Academy

16 Advanced Higher Notes (Unit ) Substituting in the initial conditions gives (check) A = and B = 1. The particular solution is thus, y = (x ) G + + Sometimes nice numbers are obtained. The constant T is essential in the (as there is a constant on the RHS of the inhomogeneous DE); try and see what happens if it is left out of the. When the form of f (x) is the same as that of a term in the CF, choose the to b f (x) ; if this is the same form as a term of the CF, try x f (x), and so on. Example 1 Find the general solution of y =. This is a slight variation of Example. The would normally be chosen as S ; however, this is of the same form as a term in the CF. So, for the try, y = S x Then, y = S S x y = 4S x 4S Substituting all this information into the non-homogeneous DE gives, (4S + 1S 16S ) x + 4S = Hence, S = 1 4 and the general solution is, y = A G e x + B 6 e x x M Patel (April 01) 16 St. Machar Academy

17 Advanced Higher Notes (Unit ) M Patel (April 01) 1 St. Machar Academy