A general solution in terms of hypergeometric functions. The general solution in terms of hypergeometric functions is y = C 1 F ( 1 , 1 , − 1 2 ; x ) + C 2 x 3 2 F ( 5 2 , 5 2 , 5 2 ; x ) _ . Formula used: Hypergeometric equation are x ( 1 − x ) y " + [ c − ( a + b + 1 ) x ] y ' − a b y = 0 here a, b and c are constants and hypergeometric function y 1 ( x ) = F ( a , b , c ; x ) for the first solution and second solution y 2 ( x ) = x 1 − c F ( a − c + 1 , b − c + 1 , 2 − c ; x ) . Calculation: The given system of equation is as 2 x ( 1 − x ) y " − ( 1 + 6 x ) y ' − 2 y = 0 . Divide by 2 in the equation 2 x ( 1 − x ) y " − ( 1 + 6 x ) y ' − 2 y = 0 . 2 x ( 1 − x ) y " 2 − ( 1 + 6 x ) y ' 2 − 2 y 2 = 0 x ( 1 − x ) y " − ( 1 2 + 3 x ) y ' − y = 0 Compare the equation x ( 1 − x ) y " − ( 1 2 + 3 x ) y ' − y = 0 with x ( 1 − x ) y " + [ c − ( a + b + 1 ) x ] y ' − a b y = 0 here c − ( a + b + 1 ) x = − ( 1 2 + 3 x ) and − a b = − 1 . In order to calculate the value of − a b , a + b + 1 and c from the equation c − ( a + b + 1 ) x = − ( 1 2 + 3 x ) . c − ( a + b + 1 ) x = − ( 1 2 + 3 x ) c − ( a + b + 1 ) x = − 1 2 − 3 x Equate the coefficient in the above equation the value of c is c = − 1 2 , and the value of a + b + 1 are shown below: ( a + b + 1 ) x = 3 x a + b + 1 = 3 On comparing the coefficients of y in equations x ( 1 − x ) y " − ( 1 2 + 3 x ) y ' − y = 0 and x ( 1 − x ) y " + [ c − ( a + b + 1 ) x ] y ' − a b y = 0 since the value of a b = 1 . Let b = 1 a is substituted in a + b + 1 = 3 as shown below: a + 1 a + 1 = 3 a 2 − 2 a + 1 = 0 Split the middle term as shown below: a 2 − a − a + 1 = 0 ( a − 1 ) 2 = 0 a = 1 , 1 Substitute the value of a = 1 in a + b + 1 = 3 . 1 + b + 1 = 3 b = 3 − 2 b = 1 Therefore, substitute the value a, b and c in hypergeometric function y 1 ( x ) = F ( a , b , c ; x ) for the first solution as y 1 ( x ) = F ( 1 , 1 , − 1 2 ; x ) and second solution y 2 ( x ) = x 1 − c F ( a − c + 1 , b − c + 1 , 2 − c ; x ) as y 2 ( x ) = x 3 2 F ( 5 2 , 5 2 , 5 2 ; x ) . The general solution is given as y = C 1 y 1 + C 2 y 2 . Substitute the value of y 1 and y 2 in y = C 1 y 1 + C 2 y 2 . y = C 1 F ( 1 , 1 , − 1 2 ; x ) + C 2 x 3 2 F ( 5 2 , 5 2 , 5 2 ; x ) Hence, the general solution in terms of hypergeometric functions is y = C 1 F ( 1 , 1 , − 1 2 ; x ) + C 2 x 3 2 F ( 5 2 , 5 2 , 5 2 ; x ) _ .

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