Explanation The differential equation can be written as: t 2 ( t + 3 ) y ″ ′ − 3 t ( t + 2 ) y ″ + 6 ( 1 + t ) y ′ − 6 y = 0 y ″ ′ − 3 ( t + 2 ) t ( t + 3 ) y ″ + 6 ( 1 + t ) t 2 ( t + 3 ) y ′ + 6 t 2 ( t + 3 ) y = 0 By comparing the differential equation with general form of differential equation ( y ″ ) ′ + p 1 ( t ) y ″ + p 2 ( t ) y ′ + p 3 ( t ) y = 0 , the value of p 1 ( t ) = − 3 ( t + 2 ) t ( t + 3 ) , p 2 ( t ) = 6 ( 1 + t ) t 2 ( t + 3 ) , and p 3 ( t ) = 6 t 2 ( t + 3 ) . Substitute, y = t 2 v ( t ) in the equation y ″ ′ − 3 ( t + 2 ) t ( t + 3 ) y ″ + 6 ( 1 + t ) t 2 ( t + 3 ) y ′ + 6 t 2 ( t + 3 ) y = 0 . t 2 v ″ ′ + ( 3 × 2 t + − 3 ( t + 2 ) t ( t + 3 ) t 2 ) v ″ + ( 3 × 2 + 2 − 3 ( t + 2 ) t ( t + 3 ) × 2 t + 6 ( t + 1 ) t 2 ( t + 3 ) t 2 ) v ′ = 0 t 2 v ″ ′ + ( 6 t 2 + 18 t − 3 t 2 − 6 t ( t + 3 ) ) v ″ + ( 6 t + 18 − 12 t − 24 + 6 + 6 t ( t + 3 ) ) v ′ = 0 t 2 v ″ ′ + ( 3 t 2 + 12 t t + 3 ) v ″ = 0 On the further simplification, the following is obtained

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Chapter 4.1, Problem 28P

To determine

To solve: The differential equation t2(t+3)y″′−3t(t+2)y″+6(1+t)y′−6y=0 with initial condition y1(t)=t2, and y2(t)=t3.

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Chapter 4.1, Problem 27P

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Elementary Differential Equations

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The differential equation t 2 ( t + 3 ) y ″ ′ − 3 t ( t + 2 ) y ″ + 6 ( 1 + t ) y ′ − 6 y = 0 with initial condition y 1 ( t ) = t 2 , and y 2 ( t ) = t 3 .

To solve: The differential equation t2(t+3)y″′−3t(t+2)y″+6(1+t)y′−6y=0 with initial condition y1(t)=t2, and y2(t)=t3.

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