13 Solving nonhomogeneous equations: Variation of the constants method

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1 13 Solving nonhomogeneous equaions: Variaion of he consans meho We are sill solving Ly = f, (1 where L is a linear ifferenial operaor wih consan coefficiens an f is a given funcion Togeher (1 is a linear nonhomogeneous ODE wih consan coefficiens, whose general soluion is, of course, y( = y h ( + y p (, where y h ( is a general soluion o he homogeneous equaion Ly = 0 an y p ( is a paricular (any soluion o (1 In he las lecure we saw how o guess y p ( by looking a he expression for f( The major rawback is, of course, he choice of f( was quie limie Here we will see how o eal wih his problem in he general case 131 Variaion of he consans for he secon orer ODE Here I woul like o sar wih he secon orer ODE y + p(y + q(y = f( (2 The general soluion o he homogeneous is given by (as we know from Lecure 11 y h ( = C 1 y 1 ( + C 2 y 2 (, where y 1 ( an y 2 ( are linearly inepenen soluions o he homogeneous equaion Now he crucial assumpion: Assume ha C 1 an C 2 are no consans, bu funcions of Tha means ha we are looking for a soluion o (2 in he form y( = C 1 (y 1 ( + C 2 (y 2 ( From now I will suppress he epenence on o simplify he arihmeics, bu you shoul remember ha C 1, C 2, y 1, y 2 are funcions of We have: y = C 1y 1 + C 2y 2 + C 1 y 1 + C 2 y 2 Here we make a myserious an somewha arbirary assumpion ha C 1y 1 + C 2y 2 = 0 (Jus a few wors: I will explain his mysery in he hir par our course, an how one can come up wih his assumpion For now a vague explanaion can be as follows: We acually nee wo coniions o eermine wo funcions C 1 ( an C 2 (, an one ODE will provie us wih only one coniion, hence we are free o choose anoher coniion, which can be aken as above Using he above, we have y = C 1y 1 + C 2y 2 + C 1 y 1 + C 2 y 2 MATH266: Inro o ODE by Arem Novozhilov, aremnovozhilov@nsueu Fall

2 Now we plug y an y ino he original equaion: Finally, we have C 1y 1 + C 2y 2 + C 1 y 1 + C 2 y 2 + p(c 1 y 1 + C 2y 2 + q(c 1 y 1 + C 2 y 2 = f = C 1 (y 1 + py 1 + qy 1 + C 2 (y 2 + py 2 + qy 2 + C 1y 1 + C 2y 2 = f = C 1y 1 + C 2y 2 = 0, C 1y 1 + C 2y 2 = f, C 1y 1 + C 2y 2 = f for wo unknown funcions C 1 (, C 2 ( Noe ha he eerminan of he sysem marix is simply he Wronskian W ( = y 1 y 2 y 2 y 1 0, because y 1 an y 2 are linearly inepenen Afer some algebra we can fin (fill in he gaps: C 1 = W 1 (y 2 f, C 2 = W 1 (y 1 f, from where by inegraion we will fin C 1 an C 2 Therefore, he general soluion for (2 is given by y( = A 1 y 1 ( + A 2 y 2 ( y 1 ( W 1 (τy 2 (τf(τ τ + y 2 ( W 1 (τy 1 (τf(τ τ, where he inegrals are evaluae wihou arbirary consans, an A 1, A 2 are new arbirary consans Example 1 Fin he general soluion o y 2y + y = e, > 0 For he homogeneous counerpar we ll fin ha y h ( = C 1 e + C 2 e, an y 1 = e, y 2 = e We ll fin he Wronskian [ e e W ( = e ] e e + e = e 2 0 Hence an C 2 = e 2 e e = 1 herefore, he general soluion is C 1 = e 2 e e = 1 = C 1 =, = C 2 = ln = ln, y( = C 1 e + C 2 e e + e ln, where I use again he convenional leers C 1 an C 2 o enoe arbirary consans The main quesion of course why i we boher o learn he meho of an eucae guess, if we have such a nice an general meho The problem here is ha very ofen for he variaion of parameers he inegrals are quie ifficul or even no possible o evaluae (see you homework problems for nice examples 2

3 132 Variaion of he consans for he n-h orer ODE There is quie sraighforwar generalizaion of he variaion of parameer meho for he case of he n-h orer equaion Ly = f In his case we have ha y h = C 1 y C n y n, where {y 1,, y n } is a funamenal se of soluions Assuming ha each C i is a funcion of one can arrive o C 1y C ny n = 0, C 1y C ny n = 0, C 1y (n C ny (n 2 n = 0, C 1y (n C ny (n 1 n = 0, which is a sysem of linear algebraic equaions wih respec o unknowns C 1,, C n Noe ha he eerminan of he marix of he sysem is he Wronskian an hence no zero The soluion can be foun using any of many available mehos (Cramer s rule, Gauss eliminaions, ec 133 Solving linear ODE wih nonconsan coefficiens This secion oes no irecly belong o his lecure an iscusses some aiional poins, which were inclue in he homework problems If you i no have any quesions solving recen homework, you can safely skip his secion Noe ha in he iscussion of he linear ODE we ha wo cases: The general heory was presene for he general linear equaions wih nonconsan coefficiens, whereas he acual soluion mehos were iscusse only for he equaions wih consan coefficiens A naural quesions o ask is wheher his fac is a simple coincience or cause by some aiional reasons An he answer is simple: in he realm of ODE equaions analyical soluions (ie, soluions ha can be wrien own wih a formula are a very rare species In he vas majoriy of cases here are no regular mehos o obain a close form soluion (which oes no mean, of course, ha here are no mehos o ackle he problem The only remarkable excepion is he linear ODE wih consan coefficiens, for which he full heory exiss This is he primary reason we concenrae mosly on he linear ODE wih consan coefficiens However, in some cases a smar subsiuion can reuce he problem a hans o a solvable one, as you shoul be aware by now from your homework problems Here is a basic example, reae in mos ODE exbooks Example 2 (Cauchy Euler equaion Cauchy Euler ODE is a linear ODE wih non-consan coefficiens of he form n y (n + n 1 a n 1 y (n a 1 y + a 0 y = 0 This equaion can be fully solve, which amouns o fining a funamenal soluion se {y 1,, y n } To show how i can be one, le me sar wih he secon orer Cauchy Euler equaion 2 y + py + qy = 0 3

4 Now I make a change of he inepenen variable = e x, hence our unknown funcion y( will be some new u(x = y(e x We nee o eermine how he erivaives will change For his recall he chain rule: We also have ha y( = u(log Hence, x f(g(x = f( x g(x, y( = u(log = where = g(x x u(x x = 1 u = u e x Similarly, 2 2 y( = u (log = x (u (xe x x = u u 2 From his we ll fin ha y = u, 2 y = u u, where he prime a y means he erivaive wih respec o an prime a u means he erivaive wih respec o x For u(x we finally obain u + (p 1u + qu = 0, which is a linear ODE wih consan coefficiens, which we solve muliple imes Similarly, i can be shown ha he n-h orer Cauchy Euler equaion can be reuce by he same change of variables o he case of a consan coefficien ODE Example 3 Consier a concree example: 3 y 2y = 6 ln, 0 This equaion afer iviing by akes he form of he Cauchy Euler equaion 2 y 2y = 6 ln The change of variables = e x bring his equaion ino he form We have u u 2u = 6xe x, u(x = y(e x u h (x = C 1 e 2x + C 2 e x, herefore a paricular soluion shoul be sough in he form u p (x = (Ax + Bxe x Plugging his expression ino he equaion, we ll fin (fill in he eails 2A 3B = 0, 6A = 6 = A = 1, B = 2 3 4

5 Therefore, an Reurning o he original variable yiels u p (x = ( x + 2/3xe x, u(x = C 1 e 2x + C 2 e x + ( x + 2/3xe x y( = u(ln = C C 2 + (2/3 ln ln Finally, I woul like o menion anoher imporan fac If we know one nonzero soluion o y + p(y + q(y = 0, we can reuce is orer by one Inee, le y 1 ( 0 be such soluion Make a subsiuion y( = y 1 (u(, where u( is a new unknown funcion Then we ge or y 1u + 2y 1u + y 1 u + p((y 1u + y 1 u + q(y 1 u = 0, y 1 u + ( 2y 1 + p(y 1 u + ( y 1 + p(y 1 + q(y 1 u = 0 Therefore, using he fac ha y 1 is a soluion an anoher change v = u, we obain which is a firs orer equaion y 1 v + ( 2y 1 + p(y 1 v = 0, 5

Mathematics in Pharmacokinetics What and Why (A second attempt to make it clearer)

Mathematics in Pharmacokinetics What and Why (A second attempt to make it clearer) Mahemaics in Pharmacokineics Wha and Why (A second aemp o make i clearer) We have used equaions for concenraion () as a funcion of ime (). We will coninue o use hese equaions since he plasma concenraions