To determine To verify: The given functions are solutions of the differential equation and determine their Wronskian. Explanation Given: The differential equation x 3 y ‴ + x 2 y ″ − 2 x y ′ + 2 y = 0 and the functions x , x 2 and 1 x . Theorem used: “If the functions p 1 , p 2 , … , p n , and g are continuous on the open interval I , then there exist exactly on solution y = ϕ ( t ) of the differential equation L [ y ] = d n y d t n + p 1 ( t ) d n − 1 y d t n − 1 + ⋯ + p n − 1 ( t ) d y d t + p n ( t ) y = g ( t ) that also satisfies the initial conditions y ( t 0 ) = y 0 , y ′ ( t 0 ) = y ′ 0 , … , y n − 1 ( t 0 ) = y 0 ( n − 1 ) , where t 0 is any point in I . this solution exists throughout the interval I .” Calculation: Consider the function y 1 = x , y 2 = x 2 and y 3 = 1 x . Now, verify that y 1 = x is the solution of the equation x 3 y ‴ + x 2 y ″ − 2 x y ′ + 2 y = 0 . Differentiate y 1 = x with respect to t . y ′ 1 = 1 Differentiate y ′ 1 = 1 with respect to t . y 1 ′ ′ = 0 Differentiate y 1 ′ ′ = 0 with respect to t . y 1 ′ ′ ′ = 0 Now substitute y 1 ′ ′ ′ , y 1 ′ ′ , y 1 ′ and y 1 in x 3 y ‴ + x 2 y ″ − 2 x y ′ + 2 y = 0 . x 3 y ‴ + x 2 y ″ − 2 x y ′ + 2 y = x 3 ( 0 ) + x 2 ( 0 ) − 2 x ( 1 ) + 2 ( x ) = − 2 x + 2 x = 0 Hence, for y 1 = x the differential equation get satisfied. Now, verify that y 2 = x 2 is the solution of the equation x 3 y ‴ + x 2 y ″ − 2 x y ′ + 2 y = 0 . Differentiate y 2 = x 2 with respect to t . y ′ 2 = 2 x Differentiate y ′ 2 = 2 x with respect to t . y 2 ′ ′ = 2 Differentiate y 2 ′ ′ = 2 with respect to t . y 2 ′ ′ ′ = 0 Now substitute y 2 ′ ′ ′ , y 2 ′ ′ , y 2 ′ and y 2 in x 3 y ‴ + x 2 y ″ − 2 x y ′ + 2 y = 0 . x 3 y ‴ + x 2 y ″ − 2 x y ′ + 2 y = x 3 ( 0 ) + x 2 ( 2 ) − 2 x ( 2 x ) + 2 ( x 2 ) = 2 x 2 − 4 x 2 + 2 x 2 = 0 Hence, for y 2 = x 2 the differential equation get satisfied

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