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# Class Note for MATH 250A with Professor Lega at UA 3

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Date Created: 02/06/15
Calculus and Differential Equations ll MATH 250 B Second order equations with constant coefficients Homogeneous equations 9 In this section we consider equations of the form dzy dy 7 b 7 0 3 dx2 dx Cy 7 where the coefficients a b and c are given constants c We need to find two linearly independent solutions y1x and y2x to the above equation 0 Try a solution of the form yx expX 0 Then must be a root of the characteristic equation aZbc0 For each equation below find two linearly independent solutions and then solve the differential equation with the given initial conditions 0 Example 1 y y 7 2y 0 with initial conditions y01YY 0 0 0 Example 2 y l 6 l 25y 0 with initial conditions y0 0 M0 1 0 Example 3 y l 2y i y 0 with initial conditions y0 1 M0 1 General Method of solution To find the general solution to an ordinary differential equation of the form ay l by l cy 0 where a7 b7 6 E R proceed as follows Find the characteristic equation a2 l b l c O and solve for the roots 1 and A2 If b2 7 436 gt 0 then the two roots are real and the general solution is y CleAlquot l CgeAQX If b2 7 436 lt O the two roots 04 i B are complex conjugate of one another with 04 E7 and B 7W The general solution is of the form y e0 C1 cos x l C2 sin x b and the 2 If b2 7 436 0 then there is a double root general solution is y C1 l C2 X eAX Non homogeneous equations 0 Recall that we only need to find one particular solution to a non homogenous linear equation 9 Such a solution may be found 0 By inspection 0 By the method of variation of parameters 0 By the method of undetermined coefficients 9 The method of variation of parameters is general and may be applied to all linear ordinary differential equations 0 The method of undetermined coefficients is specific to linear equations with constant coefficients Variation of parameters a This method assumes that we already know the general solution to the homogeneous equation Assume this solution is of the form yhx C1y1x l C2y2x where C1 and C2 are constants and y1 and y2 are two linearly independent solutions to the homogeneous equation a Look for a particular solution to the non homogeneous equation in the form ypx u1xy1 X l U2Xy2X In other words the constants are now varying o The idea is to substitute this solution into the non homogeneous equation and solve for u1x and LIQX 0 Since there are two functions to find we need a second equation which we set to u1Xy1x l u2Xy2x 0 Variation of parameters continued 0 We then have to solve the 2 x 2 system for u 1 and Hg f 7 Uiyl u zy Lib1 LIN 0 for a non homogeneous linear equation of the form 3Xy bXy CXy W o The determinant of this system which is called the Wronskian 0f Y1 afld M 7 Wow l Q Q is non zero since yl and y2 are two linearly independent solutions of the homogeneous equation Variation of parameters continued 0 As a consequence one can always solve for u 1 and Hg 0 Integration gives U1 and U2 which are then substituted back in the expression for yp to obtain a particular solution 0 Example 1 Use the method of variation of parameters to show that a particular solution to y l y 7 2y X is 1 X Yplxl 5 Z 0 Example 2 Use the method of variation of parameters to show that a particular solution to y l 6 l 25y cos4x is ypX 713 6054X gsin4xgt Method of undetermined coefficients a This method may be used to find particular solutions to linear equations with constant coefficients and with a non homogeneous term of the form fx Pmx expax cos x QmxeXpozxsin X7 where Pm and Qm are polynomials in X of degree m o The idea of the method is as follows 0 Start with a trial solution yp with the right functional form but undetermined coefficients 9 Substitute yp into the differential equation a Solve for the coefficients Method of undetermined coefficients continued The trial function yp is found as follows a If 04 i B are not roots of the characteristic equation try W Kmx expmx coswx W expmx sinwxx where Km and Lm are polynomials of degree m o If 04 i B are roots of the characteristic equation of multiplicity h then try ypx Xh Kmx expax cos x l Lmxexpozxsin x7 where Km and Lm are polynomials of degree In Method of undetermined coefficients continued 0 Example 1 Use the method of undetermined coefficients to find a particular solution to y 7 4 l 4y 12X exp2x 0 Example 2 Use the method of undetermined coefficients to find a particular solution to y l 6 l 25y cos4x a If the non homogeneous term involves linear combinations of sines and cosines with polynomial coefficients then use the the principle of superposition see next page Example 3 Use the method of undetermined coefficients to find a particular solution to y l 6 l 25y cos4x l X Principle of superposition general case 0 lfypl solves a linear equation of the form anxy an71xy 1 3100 aoXy X7 and if ypg solves the same equation with a different right hand side anxy an71xy 1 3100 aoXy 6007 then yp ypl l ypg solves anxy an1xy 1alxylaoxy f1xf2x 0 As a consequence to find a particular solution yp to the last equation it may be easier to solve for the first two equations separately and then write yp as the sum of ypl and ypg

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