Explanation The given differential equation is t 2 y ″ + 3 t y ′ + y = 0 , t > 0 . Compare t 2 y ″ + 3 t y ′ + y = 0 with general form t 2 y ″ + α t y ′ + β y = 0 . Here, α = 3 and β = 1 . Substitute α = 3 and β = 1 in d 2 y d t 2 + ( α − 1 ) d y d t + β y = 0 . d 2 y d t 2 + ( 3 − 1 ) d y d t + ( 1 ) y = 0 d 2 y d t 2 + 2 d y d t + y = 0 y ″ + 2 y ′ + y = 0 Obtain the characteristic equation by replacing y ″ with r 2 , y ′ with r and y with 1 in the differential equation: r 2 + 2 r + 1 = 0 . The roots of the characteristic equation are as follows. r 2 + 2 r + 1 = 0 r 1 , 2 = − 2 ± ( 2 ) 2 − 4 ( 1 ) ( 1 ) 2 ( 1 ) ( ∵ x 1 , 2 = − b ± b 2 − 4 a c 2 a ) r 1 , 2 = − 2 ± 4 − 4 2 r 1 , 2 = − 2 ± 0 2 Simplify further, r 1 , 2 = − 2 2 r = − 1 Note that, the general equation is y ( t ) = c 1 e r t + c 2 t e r t where r is the double root of the characteristic equation

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Chapter 3.4, Problem 43P

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The solution of the given differential equation using the substitution introduced problem 34 in Section 3.3.

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