Soln_Ass4_AMATH351_topost - AM 351 Assignment 4 Solutions 1...

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AM 351: Assignment 4: Solutions 1. a) We look for a solution of the form x r and obtain the characteristic equation, r 2 + ( α - 1) r + β = 0 . The roots of which are, r = 1 - α ± p ( α - 1) 2 - 4 β 2 . If we only have one root then that requires ( α - 1) 2 = 4 β . But from this substitute we find one solution, y 1 ( x ) = x r 1 , where r 1 = (1 - α ) / 2 or α = 1 - 2 r 1 . We look for a second solution of the form y 2 = v ( x ) y 1 and know, from our previous work that v is determined by the following equation, v 0 = 1 x 2 r 1 exp ± - Z α x dx ² = 1 x 2 r 1 exp ( - α ln x ) = x 2 r 1 - 1 x 2 r 1 = 1 x . Therefore, v = ln x and y 2 = x r 1 ln x b) From assignment 1 recall that if we substitute x = e t then we get dy dx = e - t dy dt , d 2 y dx 2 = e - 2 t ³ d 2 y dt 2 - dy dt ´ . Similarly, we obtain for the third order derivative, d 3 y dx 3 = e - 3 t ³ d 3 y dt 3 - 3 d 2 y dt 2 + 2 dy dt ´ When we substitute this into the equation we get, d 3 y dt 3 - 3 d 2 y dt 2 + 3 dy dt - y = 0 . If we try y = e λt then the characteristic equation is, λ 3 - 3 λ 2 + 3 λ - 1 = 0 This is a cubic equation that can be very difficult to solve but observe that λ = 1 is a valid root. Then by long division we get the other two and ( λ - 1) 3 = 0 Therefore the roots are λ = 1 with multiplicity of 3. We know that the general solution to this equation should be, y ( t ) = c 1 e t + c 2 te t + c 3 t 2 e t , and when we substitute back in terms of x we get, y ( x ) = c 1 x + c 2 (ln x ) x + c 3 (ln x ) 2 x. 1
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2. a) We compute p 0 and q 0 , as defined in lecture, p 0 = lim x 1 ( x - 1) ± - 2 x 1 - x 2 ² = 2 2 = 1 , q 0 = lim x 1 ( x - 1) 2 ± p ( p + 1) 1 - x 2 ² = 0 . So the associated Euler equation is, x 2 y 00 + xy 0 = 0 . The characteristic equation is r ( r - 1) + r = 0 that implies r = 0 is a double root. b) Before we look for a power series solution I choose to substitute t = x - 1. If we rewrite Y ( t ) = y ( x ) then the equation is, ( - t 2 - 2 t ) Y 00 + ( - 2 t - 2) Y 0 + p ( p + 1) Y = 0 . Therefore, we look for a regular power series solution of the form, Y = X n =0 a n t n We substitute this into the equation and rewrite each term separately. p ( p + 1) Y = X n =0 p ( p + 1) a n t n , - 2 tY 0 = X n =0 ( - 2 n ) a n t n , - 2 Y 0 = X n =1 ( - 2 n ) a n t n - 1 = X n =0 ( - 2( n + 1)) a n +1 t n , - t 2 Y 00 = X n =0 - n ( n - 1) a n t n , - 2 tY 00 = X n =0 - 2 n ( n - 1) a n t n - 1 = X n =0 - 2( n + 1)( n ) a n +1 t n .
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Soln_Ass4_AMATH351_topost - AM 351 Assignment 4 Solutions 1...

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