Explanation The given first order differential equation is, s n = a s n − 1 + b . First substitute n = 1 in s n = a s n − 1 + b and obtain s 1 as follows. s 1 = a s 1 − 1 + b = a s 0 + b Now using the result s n = { a n ( s 0 + c ) − c if a ≠ 1 s 0 + n b if a = 1 , where c = b a − 1 obtain s 1 for a ≠ 1 as shown below. s 1 = a 1 ( s 0 + c ) − c = a s 0 + a c − c = a s 0 + a ( b a − 1 ) − ( b a − 1 ) = a s 0 + b ( a − 1 a − 1 ) = a s 0 + b For a = 1 obtain s 1 as follows. s 1 = s 0 + ( 1 ) b = s 0 + b Hence the result s n = { a n ( s 0 + c ) − c if a ≠ 1 s 0 + n b if a = 1 , where c = b a − 1 is correct for n = 1 . Now assume the formula is correct for n = 2 , 3 , … , k . Then by assumption, s k = { a k ( s 0 + c ) − c if a ≠ 1 s 0 + k b if a = 1 Prove the formula is correct for n = k + 1 as follows. For a ≠ 1 obtain s k + 1 as shown below. s k + 1 = a s k + 1 − 1 + b = a s k + b = a ( a k ( s 0 + c ) − c ) + b = a k + 1 ( s 0 + c ) − a c + b Substitute c = b a − 1 and simplify further as shown below

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Author: Dossey, John A.

Publisher: Pearson,

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Chapter 9.3, Problem 35E

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To prove: The term sn={an(s0+c)−c if a≠1s0+nb if a=1 where c=ba−1 is a solution to the first order differential equation sn=asn−1+b by mathematical induction.

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Chapter 9.3, Problem 34E

Chapter 9.3, Problem 36E

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To prove: The term s n = { a n ( s 0 + c ) − c if a ≠ 1 s 0 + n b if a = 1 where c = b a − 1 is a solution to the first order differential equation s n = a s n − 1 + b by mathematical induction.

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To prove: The term sn={an(s0+c)−c if a≠1s0+nb if a=1 where c=ba−1 is a solution to the first order differential equation sn=asn−1+b by mathematical induction.

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Chapter 9 Solutions