notes1

notes1 - is not a constant hence y 1 = e rx and y 2 = u x e...

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MATH 1005A - Notes 1 Reduction of Order Consider a linear, second-order, homogeneous equation in standard form, y 00 + p ( x ) y 0 + q ( x ) y 0 =0 : Suppose that one solution y 1 is known. Then a second, independent solution y 2 is obtained by letting y 2 ( x )= u ( x ) y 1 ( x ) ;u ( x ) to be determined. y 2 = uy 1 ) y 0 2 = u 0 y 1 + uy 0 1 ;y 00 = u 00 y 1 +2 u 0 y 0 1 + uy 00 1 ; so y 2 is a solution if and only if [ u 00 y 1 +2 u 0 y 0 1 + uy 00 1 ]+ p ( x )[ u 0 y 1 + uy 0 1 ]+ q ( x )[ uy 1 ]=0 ; i.e., u [ y 00 1 + p ( x ) y 0 1 + q ( x ) y 1 ]+ u 00 y 1 +2 u 0 y 0 1 + p ( x ) u 0 y 1 =0 : Since y 1 is a solution, y 00 1 + p ( x ) y 0 1 + q ( x ) y 1 =0. Thus , y 2 is a solution if and only if u 00 y 1 + u 0 [2 y 0 1 + p ( x ) y 1 ]=0 ; i.e., u 00 u 0 = ¡ 2 y 0 1 + p ( x ) y 1 y 1 = ¡ 2 y 0 1 y 1 ¡ p ( x ) : Integrat ionw ithrespectto x then gives ln j u 0 j = ¡ 2ln j y 1 Z p ( x ) dx: Taking the exponential of both sides and using the fact that e ¡ 2ln j y 1 j = e ln j y 1 j ¡ 2 = 1 y 2 1 ,w e obtain j u 0 j = 1 y 2 1 e ¡ R p ( x ) dx ; or u 0 = § 1 y 2 1 e ¡ R p ( x ) dx : Taking the plus sign and integrating once more, we obtain u ( x )= Z 1 y 2 1 e ¡ R p ( x ) dx dx: Since 1 y 2 1 e ¡ R p ( x ) dx 6 =0 ;

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Unformatted text preview: ) is not a constant, hence y 1 = e rx and y 2 = u ( x ) e rx are linearly independent. For the equation ay 00 + by + cy = 0 with b 2 Â¡ 4 ac = 0 ; y = e rx ) ar 2 + br + c = 0 ) r = Â¡ b Â§ p b 2 Â¡ 4 ac 2 a = Â¡ b 2 a ) y 1 = e rx = e Â¡ b 2 a x : 2 In standard form, the equation is y 00 + b a y + c a y = 0 ; with p ( x ) = b a : Thus, by reduction of order, u = 1 y 2 1 e Â¡ R p ( x ) dx = 1 e 2 rx e Â¡ R b a dx = 1 e 2 rx e Â¡ b a x = 1 e 2 rx e 2 rx = 1 ) u ( x ) = x ) y 2 = xe rx :...
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notes1 - is not a constant hence y 1 = e rx and y 2 = u x e...

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