A general solution in terms of hypergeometric functions. The general solution in terms of hypergeometric functions is y = C 1 F ( − 2 , − 1 , 1 2 ; x ) + C 2 x 1 2 F ( − 3 2 , − 1 2 , 3 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 x ( 1 − x ) y " + ( 1 2 + 2 x ) y ' − 2 y = 0 . Compare the equation x ( 1 − x ) y " + ( 1 2 + 2 x ) y ' − 2 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 + 2 x ) and − a b = − 2 . In order to calculate the value of − a b , a + b + 1 and c from the equation c − ( a + b + 1 ) x = ( 1 2 + 2 x ) . c − ( a + b + 1 ) x = ( 1 2 + 2 x ) c − ( a + b + 1 ) x = 1 2 + 2 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 = − 2 x a + b + 1 = − 2 On comparing the coefficients of y in equations x ( 1 − x ) y " + ( 1 2 + 2 x ) y ' − 2 y = 0 and x ( 1 − x ) y " + [ c − ( a + b + 1 ) x ] y ' − a b y = 0 since the value of a b = 2 . Let b = 2 a is substituted in a + b + 1 = − 2 as shown below: a + 2 a + 1 = − 2 a 2 + 3 a + 2 = 0 By splitting the middle term as shown below: a 2 + 2 a + a + 2 = 0 ( a + 1 ) ( a + 2 ) = 0 a = − 1 , − 2 Substitute the value of a = − 2 in a + b + 1 = − 2 . − 2 + b + 1 = − 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 ( − 2 , − 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 1 2 F ( − 3 2 , − 1 2 , 3 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 ( − 2 , − 1 , 1 2 ; x ) + C 2 x 1 2 F ( − 3 2 , − 1 2 , 3 2 ; x ) Hence, the general solution in terms of hypergeometric functions is y = C 1 F ( − 2 , − 1 , 1 2 ; x ) + C 2 x 1 2 F ( − 3 2 , − 1 2 , 3 2 ; x ) _ .

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Solve the initial value problem: y' = x3 + y³ 3 , y(1) = 2 xy² y(x) = Hint: Notice that the equation on the right is homogeneous and see Homework exercise 23 in section 1.2 of our textbook to review techniques for solving homogeneous equations. Note that we've been given an intial value of the form y(a) = b where a > 0, so this only determines a solution corresponding to the right half of the graph of In(x), i.e., the part of the graph corresponding to positive values of x. Therefore, we should write In(x) instead of ln(|x|), since the left half of the graph is not determined by the initial condition given.