A general solution in terms of hypergeometric functions. The solution of the ordinary differential equation is y = A 1 F ( 0.33 , − 1 , 0.33 ; t + 1 ) + A 2 ( t + 1 ) 0.66 F ( − 0.33 , 1 , 0.67 ; t + 1 ) _ . 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 1 + D = − c ( t 2 − t 1 ) , C = a + b + 1 and K = a b . The value of F ( a , b , c ; x ) is equal to F ( b , a , c ; x ) , does not affect the solution of ordinary differential equation. Calculation: The given system of equation is as 3 t ( 1 + t ) y ″ + t y ′ − y = 0 or t ( 1 + t ) y ″ + 0 . 3 3 t y ′ − 0 . 3 3 y = 0 Now for the value of x factorized the coefficient of y ″ . t ( 1 + t ) = 0 t 1 = − 1 t 2 = 0 Substitute the value of t 1 and t 2 in x = t − t 1 t 2 − t 1 . x = t − ( − 1 ) 0 − ( − 1 ) = t + 1 1 = t + 1 When compare the given second order equation t ( 1 + t ) y ″ + 0 . 3 3 t y ′ − 0 . 3 3 y = 0 with ( t 2 + A t + B ) y ″ + ( C t + D ) y ′ + K y = 0 then the value C = 0.33 , D = 0 and K = − 0.33 . Now substitute the value C = 0.33 in equation C = a + b + 1 . C = a + b + 1 a + b + 1 = 0.33 a + b = − 0.66 a = − 0.66 − b Substitute, the value a = − 0.66 − b in a b = − 0.33 . ( − 0.66 − b ) b = − 0.33 − 0.66 b − b 2 = − 0.33 b 2 + 0.66 b − 0.33 = 0 Further simplification of above equation. b = − 0.66 ± 0.66 2 − 4 × 1 × ( − 0.33 ) 2 × 1 = − 0.66 ± 1.98 2 × 1 = − 0.66 ± 1.40 2 = 0.37 , − 1.03 Now substitute the value b = 0.37 in a = − 0.66 − b . a = − 0.66 − 0.37 = − 1.03 ≃ − 1 So, the value of a = − 1 , b = 0.37 or b = − 1 , a = 0.37 . Because the function F ( a , b , c ; x ) = F ( b , a , c ; x ) . 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 ) . 0.33 ( − 1 ) + 0 = − c ( 0 + 1 ) c = 0.33 Substitute the all coefficient value in general ordinary differential equation x ( 1 − x ) y ″ + [ c − ( a + b + 1 ) x ] y ′ − a b y = 0 . x ( 1 − x ) y ″ + [ 0.33 − ( − 0.66 + 1 ) x ] y ′ − ( − 0.33 ) y = 0 x ( 1 − x ) y ″ + [ 0.33 − 0.33 x ] y ′ + ( 0.33 ) 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 ( b , a , c ; x ) + A 2 x 1 − c F ( a − c + 1 , b − c + 1 , 2 − c ; x ) = A 1 F ( 0.33 , − 1 , 0.33 ; t + 1 ) + A 2 x 1 − 0.33 F ( − 1 − 0.33 + 1 , 0.33 − ( 0.33 ) + 1 , 2 − ( 0.33 ) ; t + 1 ) = A 1 F ( 0.33 , − 1 , 0.33 ; t + 1 ) + A 2 x 0.66 F ( − 0.33 , 1 , 0.67 ; t + 1 ) Therefore, the solution of the ordinary differential equation is y = A 1 F ( 0.33 , − 1 , 0.33 ; t + 1 ) + A 2 ( t + 1 ) 0.66 F ( − 0.33 , 1 , 0.67 ; t + 1 ) _ .

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