A general solution in terms of hypergeometric functions. The solution of the ordinary differential equation is y = A 1 F ( − 2 , 2 , − 0.5 ; t − 2 ) + A 2 x 1.5 F ( − 0.5 , 3.5 , 2.5 ; t − 2 ) _ . Formula used: The general solution of hypergeometric equation ( t 2 + A t + B ) y ″ + ( C t + D ) y ′ + K y = 0 is y = A 1 y 1 + A 2 y 2 . Where y 1 = F ( a , b , c ; x ) , y 2 = x 1 − c F ( a − c + 1 , b − c + 1 , 2 − c ; x ) and A, a, B, b, C, c, D, K are constant. Where the value x = t − t 1 t 2 − t 1 and parameter related by C t + D = − c ( t 2 − t 1 ) , C = a + b + 1 and K = a b . Calculation: The given system of equation is as 2 ( t 2 − 5 t + 6 ) y ″ − ( 2 t − 3 ) y ′ − 8 y = 0 or ( t 2 − 5 t + 6 ) y ″ − ( t − 1.5 ) y ′ − 4 y = 0 Now for the value of x factorized the coefficient of y ″ . 2 ( t 2 − 5 t + 6 ) = 0 t 2 − 2 t − 3 t + 6 = 0 t ( t − 2 ) − 3 ( t − 2 ) = 0 Further simplification of above equation. ( t − 2 ) ( t − 3 ) = 0 t 1 = 2 t 2 = 3 Substitute the value of t 1 and t 2 in x = t − t 1 t 2 − t 1 . x = t − 2 3 − 2 = t − 2 1 = t − 2 When compare the given second order equation 2 ( t 2 − 5 t + 6 ) y ″ − ( 2 t − 3 ) y ′ − 8 y = 0 with ( t 2 + A t + B ) y ″ + ( C t + D ) y ′ + K y = 0 then the value C = 1 , D = − 1.5 and K = − 4 . Now substitute the value C = 1 in equation C = a + b + 1 . C = a + b + 1 a + b + 1 = 1 a + b = 0 a = − ( b ) Substitute, the value a = − ( b ) in a b = − 4 . ( − b ) ( b ) = − 4 b 2 = 4 b = 2 Now substitute the value b = 2 in a = − b . a = − 2 For the value of c , substitute the value of C , D and ( t 2 − t 1 ) in equation C t 1 + D = − c ( t 2 − t 1 ) . 1 × 2 − 1.5 = − c ( t 2 − t 1 ) c ( 3 − 2 ) = − 0.5 c = − 0.5 Substitute the all coefficient value in general ordinary differential equation x ( x − 1 ) y ″ + [ c − ( a + b + 1 ) x ] y ′ − a b y = 0 . x ( x − 1 ) y ″ + [ − 0.5 − ( − 2 + 2 + 1 ) x ] y ′ + 4 y = 0 x ( x − 1 ) y ″ + [ − 0.5 − ( 0 + 1 ) x ] y ′ + 4 y = 0 x ( x − 1 ) y ″ − ( 0.5 + x ) y ′ + 4 y = 0 The obtained equation is ordinary differential equation, so, the general solution of equation is calculated as: y = A 1 y 1 + A 2 y 2 = A 1 F ( a , b , c ; x ) + A 2 x 1 − c F ( a − c + 1 , b − c + 1 , 2 − c ; x ) = A 1 F ( − 2 , 2 , − 0.5 ; t − 2 ) + A 2 x 1 + 0.5 F ( − 2 + 0.5 + 1 , 2 + 0.5 + 1 , 2 + 0.5 ; t − 2 ) = A 1 F ( − 2 , 2 , − 0.5 ; t − 2 ) + A 2 x 1.5 F ( − 0.5 , 3.5 , 2.5 ; t − 2 ) Therefore, the solution of the ordinary differential equation is y = A 1 F ( − 2 , 2 , − 0.5 ; t − 2 ) + A 2 x 1.5 F ( − 0.5 , 3.5 , 2.5 ; t − 2 ) _ .

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