A basis of solutions by the Frobenius method in detail. The basis of solutions by the Frobenius method y = c 1 e x − c 2 e x x _ . Theorem used: Let r 1 and r 2 be the roots of the indicial equation. Then the following three cases as: Case 1: Distinct roots not differing by an integer. A basis is y 1 ( x ) = x r 1 ( a 0 + a 1 x + a 2 x 2 + ⋅ ⋅ ⋅ ) and y 2 ( x ) = x r 2 ( A 0 + A 1 x + A 2 x 2 + ⋅ ⋅ ⋅ ) with coefficients obtained successively with r = r 1 and r = r 2 respectively. Case 2: Double roots r 1 = r 2 = r . A basis is y 1 ( x ) = x r 1 ( a 0 + a 1 x + a 2 x 2 + ⋅ ⋅ ⋅ ) and y 2 ( x ) = y 1 ( x ) ln x + x r ( A 1 x + A 2 x 2 + ⋅ ⋅ ⋅ ) . Case 3: Roots differing by an integer. A basis is y 1 ( x ) = x r 1 ( a 0 + a 1 x + a 2 x 2 + ⋅ ⋅ ⋅ ) and y 2 ( x ) = k y 1 ( x ) ln x + x r 2 ( A 0 + A 1 x + A 2 x 2 + ⋅ ⋅ ⋅ ) where the roots are so denoted that r 1 − r 2 > 0 and k may turn out to be zero. Calculation: The given system of equation is as x y " + ( 2 − 2 x ) y ' + ( x − 2 ) y = 0 . Compare the equation x y " + ( 2 − 2 x ) y ' + ( x − 2 ) y = 0 with P ( x ) y " + Q ( x ) y ' + R ( x ) y = 0 here P ( x ) = x , Q ( x ) = 2 − 2 x and R ( x ) = x − 2 are all ordinary points. Let a series solution in powers of x as y ( x ) = ∑ m = 0 ∞ a m x m + r . Differentiate the equation y ( x ) = ∑ m = 0 ∞ a m x m + r as shown below: d d x ( y ) = d d x ( ∑ m = 0 ∞ a m x m + r ) y ' = ∑ m = 0 ∞ ( m + r ) a m x m + r − 1 Again, differentiate the equation y ' = ∑ m = 0 ∞ ( m + r ) a m x m + r − 1 as shown below: d d x ( y ) = d d x ( ∑ m = 0 ∞ ( m + r ) a m x m + r − 1 ) y " = ∑ m = 0 ∞ ( m + r ) ( m + r − 1 ) a m x m + r − 2 Substitute the value of y , y ' and y " in x y " + ( 2 − 2 x ) y ' + ( x − 2 ) y = 0 . x ( ∑ m = 0 ∞ ( m + r ) ( m + r − 1 ) a m x m + r − 2 ) + ( 2 − 2 x ) ( ∑ m = 0 ∞ ( m + r ) a m x m + r − 1 ) + ( x − 2 ) ( ∑ m = 0 ∞ a m x m + r ) = 0 a m [ x ( m + r ) ( m + r − 1 ) a m x m + r − 2 + ( 2 − 2 x ) ( m + r ) x m + r − 1 + ( x − 2 ) x m + r ] = 0 Therefore, the lowest power is x m + r − 1 and the corresponding equation is shown below: [ r ( r − 1 ) + 2 r ] a m = 0 r ( r − 1 ) + 2 r = 0 r 2 − r + 2 r = 0 r 2 + r = 0 Further simplification as shown below: r 2 + r = 0 r ( r + 1 ) = 0 r = 0 , − 1 Hence the roots of the above equation are r 1 = 0 and r 2 = − 1 . In order to calculate the first solution, equalize the power of x by filling in m = ( n + 1 ) at x m + r − 1 , m = n at x m + r and m = ( n − 1 ) at x m + r + 1 and further substitute r 1 = 0 in x ( ∑ m = 0 ∞ ( m + r ) ( m + r − 1 ) a m x m + r − 2 ) + ( 2 − 2 x ) ( ∑ m = 0 ∞ ( m + r ) a m x m + r − 1 ) + ( x − 2 ) ( ∑ m = 0 ∞ a m x m + r ) = 0 x n [ n ( n + 1 ) a n + 1 + 2 ( n + 1 ) a n + 1 − 2 n a n + a n − 1 − 2 a n ] = 0 a n + 1 [ n 2 + n + 2 n + 2 ] + a n [ − 2 n − 2 ] + a n − 1 = 0 a n + 1 [ n 2 + 3 n + 2 ] − 2 n a n − 2 a n + a n − 1 = 0 a n + 1 = ( 2 n + 2 ) a n − a n − 1 n 2 + 3 n + 2 Further simplification as shown below: a n + 1 = ( 2 n + 2 ) a n − a n − 1 n 2 + 3 n + 2 a n + 1 = ( n + 1 ) 2 a n − a n − 1 ( n + 1 ) ( n + 2 ) a n + 1 = 2 a n ( n + 2 ) − a n − 1 ( n + 1 ) ( n + 2 ) Substitute k = ( n + 1 ) in the equation a n + 1 = 2 a n ( n + 2 ) − a n − 1 ( n + 1 ) ( n + 2 ) . a k = 2 a k − 1 ( k + 1 ) − a k − 2 k ( k + 1 ) On choosing a 0 = 1 , therefore for the first several values of a n , the above equation as shown below: a 1 = 2 a 1 − 1 ( 1 + 1 ) − a 1 − 2 1 ( 1 + 1 ) = 2 a 0 2 − a − 1 2 Substitute the value of a 0 = 1 and a − n = 0 in the above equation. a 1 = 2 ( 1 ) 2 − ( 0 ) 2 = 2 2 = 1 a 2 = 2 a 2 − 1 ( 2 + 1 ) − a 2 − 2 2 ( 2 + 1 ) = 2 a 1 3 − a 0 6 Substitute the value of a 0 = 1 and a 1 = 1 a 2 = 2 3 × 2 2 − ( 1 ) 6 = 2 3 × 2 2 − 1 6 = 3 2 × 3 a n = 1 n ! Thus, the solution has the form as shown below: y 1 ( x ) = ∑ n = 1 ∞ a n x n + r Substitute the value of r 1 = 0 and a n = 1 n ! in the equation y 1 ( x ) = ∑ n = 1 ∞ a n x n + r . y 1 ( x ) = ∑ n = 1 ∞ 1 n ! x n Hence the above equation is series from the Maclaurin series, since the above equation becomes as shown below: y 1 ( x ) = e x Hence the first solution of the Frobenius method is y 1 ( x ) = e x _ . In order to calculate the second function on applying reduction of order. Consider the value y 2 = u y 1 . Form the reduction of order of ODE the derivative of U = 1 y 1 2 e − ∫ p d x here U = u ' . From the given differential equation x y " + ( 2 − 2 x ) y ' + ( x − 2 ) y = 0 , the value of p ( x ) is calculated as: y " + ( 2 − 2 x ) x y ' + ( x − 2 ) x y = 0 Now the compare the above equation with the general equation y " + p ( x ) y ' + q ( x ) y = r ( x ) here p ( x ) = ( 2 − 2 x ) x . Substitute the value of p ( x ) = ( 2 − 2 x ) x in − ∫ p ( x ) d x . − ∫ p ( x ) d x = − ∫ ( 2 − 2 x ) x d x = ∫ − 2 x + 2 d x = − 2 ln ( x ) + 2 x = ln ( x − 2 ) + 2 x Substitute the value of y 1 ( x ) = e x and − ∫ p ( x ) d x = ln ( x − 2 ) + 2 x in u ' = 1 y 1 2 e − ∫ p d x . u ' = 1 ( e x ) 2 e ln ( x − 2 ) + 2 x = x − 2 e 2 x e 2 x = x − 2 Therefore, the value of u from the above equation is, u = ∫ x − 2 d x = − 1 x Substitute the value of u = − 1 x and y 1 ( x ) = e x in y 2 = u y 1 . y 2 = − 1 x e x = − e x x The general solution is 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 e x − c 2 e x x Hence, the basis of solutions by the Frobenius method y = c 1 e x − c 2 e x x _ .

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