Explanation Procedure used: Consider the differential equation a t 2 y ″ ( t ) + b t y ′ ( t ) + c y = 0 . The associated auxiliary equation is a r 2 + ( b − a ) r + c = 0 . Solving the auxiliary equation gives the roots r 1 and r 2 . (a) If the auxiliary equation has distinct real roots r 1 and r 2 , then y ( t ) = c 1 t r 1 + c 2 t r 2 is a general solution. (b) If the auxiliary equation has repeated root r , then y ( t ) = c 1 t r + c 2 t r ln t is a general solution. (b) If the auxiliary equation has complex root r = α ± i β , then y ( t ) = c 1 t α cos ( β ln t ) + c 2 t α sin ( β ln t ) is a general solution. Calculation: The given initial value problem is, t 2 y ″ ( t ) − 4 t y ′ ( t ) + 4 y ( t ) = 0 ; y ( 1 ) = − 2 , y ′ ( 1 ) = − 11 . Comparing the differential equation t 2 y ″ ( t ) − 4 t y ′ ( t ) + 4 y ( t ) = 0 with the general form a t 2 y ″ ( t ) + b t y ′ ( t ) + c y = 0 gives a = 1 , b = − 4 and c = 4 . The associated auxiliary equation is, a r 2 + ( b − a ) r + c = 0 becomes, 1 r 2 + ( − 4 − 1 ) r + ( 4 ) = 0 r 2 − 5 r + 4 = 0 Factorize the polynomial r 2 − 5 r + 4 = 0 as, r 2 − 4 r − r + 4 = 0 r ( r − 4 ) − 1 ( r − 4 ) = 0 ( r − 4 ) ( r − 1 ) = 0 r = 1 , 4 Note that, the roots r 1 = 1 and r 2 = 4 are distinct and real. Thus, the given differential equation has solution of the form y ( t ) = c 1 t + c 2 t 4

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Chapter 4.7, Problem 19E

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The general solution to the initial value problem t2y″(t)−4ty′(t)+4y(t)=0;y(1)=−2,y′(1)=−11.

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The general solution to the initial value problem t 2 y ″ ( t ) − 4 t y ′ ( t ) + 4 y ( t ) = 0 ; y ( 1 ) = − 2 , y ′ ( 1 ) = − 11 .

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The general solution to the initial value problem t2y″(t)−4ty′(t)+4y(t)=0;y(1)=−2,y′(1)=−11.

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