Physical Chemistry Introduction to Quantum Mechanics
Physical Chemistry Introduction to Quantum Mechanics CHEM 113A
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This 3 page Class Notes was uploaded by Michael Reilly on Friday September 4, 2015. The Class Notes belongs to CHEM 113A at University of California - Los Angeles taught by Staff in Fall. Since its upload, it has received 122 views. For similar materials see /class/177989/chem-113a-university-of-california-los-angeles in Chemistry at University of California - Los Angeles.
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Date Created: 09/04/15
BRIEF REVIEW OF CONSTANT COEFFICIENT SECONDORDER LINEAR DIFFERENTIAL EQUATIONS This review follows 77Calculus77 by Stewart Edition 4 Chapter 18 First we review some general facts about secondorder linear differential equations A secondorder linear differential equation SOLDE has the form d2y dy PltacgtE exam Rltzgty i am lt1 where P Q R and Q are continuous functions If Cz 0 in 1 then de dy PIQIERIyio 2 and the SOLDE is called homogeneous if Cz y 0 it is called nonhomo geneousi FACT 1 If y1z and yg are solutions to the homogeneous SOLDE of 2 then so is any linear combination of y1 and yg iiei 91 01911 C2y2I is also a solution to the homogeneous SOLDE in 2 for any constants cl and 02 Two functions y1x and y2x are called linearly independant if neither yl or yg is a constant multiple of the other yl cyg for c f 0 FACT 2 If yl and yg are linearly independant solutions to 2 and Pz is never 0 then every solution to 2 can be written as a linear combination of y1 and y2z ie the general solution to 2 is given by 91 01911 02929 where cl and Cg are arbitrary constantsi CONSTANT COEFFICIENT SECONDORDER LINEAR DIFFERENTIAL EQUATIONS What we need so far in our MATH 3131 class is just knowledge on how to solve constant coef cient SOLDEs7 iiei7 equations of the form W by Cy 0 3 where ab and c are constants7 a f 0 and denotes it In fact7 all we need so far in Haberman is to solve 3 for a l and b 0 just try to remember that as you read below7 and compare with class lecture notes If we put y 6 into 37 then we get em a39r2 W c 07 and since 6 f 07 then aT2bTc0i 4 4 is called the characteristic equation which is just a quadratic equation7 which has the two roots 71 V122 7 4ac 7 7 V122 7 4ac 5 7 1 7 2 2a 2a We look at 3 cases which depend on the discriminant 2 7 4ac where its gt00andlt0i Case 1 b2 7 4ac gt 0 In this case7 there are two real roots 7 1 and T2 and so the general solution is y 616nm 026 In the special case where a 17 b 0 and c lt 07 then 745 T13 iT i7 and so the general solution can be written as y 5167 626771 6 But since the hyperbolic sine and cosine are we 7 771 we 771 sinhyz coshyz i and e sinhyz cosheyz7 6 7 7sinhyz cosheyz7 then we can write the general solution 6 in the case of a 17 b 0 and c lt 0 as a linear combination of sinhyz and coshyz y d1 sinhWI d2 COSh I for some constants d1 and d2 and 7 gr We note heTe that this is the case that the class was having tTouble undeT standing namely the equation 0172 7Aq foT A lt 0 In the above notation y 45 c A Example 1 Solve y 7 3y 0 Solution Put y em The characteristic equation is T273 0 Therefore the roots are T13 i i The general solution Will be y d1 sinhz d2 coshz Case 2 b2 7 4ac 0 In this case there is only one real root T 7b2a and therefore e is a solution The other independant solution Will be me check that this satis es the SOLDE 3 if e is a solution Therefore the general solution Will be y Clem ogre Example 2 Solve y 7 4y 4 0 Solution The characteristic equation is 7 2 7 4T T 7 22 0 Therefore there is only one root T 2 and the general solution to y 7 4y 4 0 is y cleh ngeh Case 3 b2 7 4ac lt 0 In this case there are two complex roots T1ai T2a7i Where a 7b2a and WZa But by Eulerls equation we have em cost9 isint9 and therefore the general solution can be Written as y Clem C2620 Cleo i x 624047113 Clem cos 61 i sin age cos Br 7 isin EMEKCl W 00551 i51 C2Sin51l em C1 cos 61 C2 sin