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 .

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Chapter 4.1, Problem 28P

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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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