SOLDE_1-11

SOLDE_1-11 - Tutorial SOLDE: Second-Order Linear...

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Tutorial SOLDE: Second-Order Linear Differential Equations 5. Solutions to Exercises Exercise 1 For y ( x ) = ® 1 y 1 ( x ) + ® 2 y 2 ( x ) , y 0 ( x ) = ® 1 y 0 1 ( x ) + ® 2 y 0 2 ( x ) and y 00 ( x ) = ® 1 y 00 1 ( x )+ ® 2 y 00 2 ( x ) : Substituting into Eqn. 2 = ) ® 1 [ ay 00 1 + by 0 1 + cy 1 ]+ ® 2 [ ay 00 2 + by 0 2 + cy 2 ] = ® 1 £ 0 + ® 2 £ 0 =0 : Exercise 2 d dx ( y 1 y 0 2 ¡ y 2 y 0 1 ) = y 0 1 y 0 2 + y 1 y 00 2 ¡ y 0 2 y 0 1 ¡ y 2 y 00 1 = y 1 y 00 2 ¡ y 2 y 00 1 : The rest of the derivation of Eqn. 5 involves dividing by a ( x ) : Exercise 3 If y 2 = ®y 1 ; then y 0 2 = ®y 0 1 ; and Eqn. 5 becomes d=dx (0) + P ( x )(0) = 0 ; or 0 =0 : This is a trivial equation from which no further information can be obtained. Exercise 4 We assume that W ( y 1 ; y 2 ) =0 : Then, y 0 2 y 2 = y 0 1 y 1 = ) ln y 2 = ln y 1 +ln C = ) y 2 = C 0 y 1 ; where, in obtaining the second expression, we found the antiderivative of the first. The vanishing of the Wronskian therefore implies the proportionality of the two functions. Exercise5 The differences in thecoefficients C ij are therebecause thesecoefficientscandepend numerically on the exact form of the two solutions involved. The demonstration involves only simple algebraic steps:
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SOLDE_1-11 - Tutorial SOLDE: Second-Order Linear...

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