A general solution in terms of hypergeometric functions. The solution of the ordinary differential equation is y = A 1 F ( − 2 , 2 , − 0.5 ; t − 2 ) + A 2 x 1.5 F ( − 0.5 , 3.5 , 2.5 ; t − 2 ) _ . 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 + D = − c ( t 2 − t 1 ) , C = a + b + 1 and K = a b . Calculation: The given system of equation is as 2 ( t 2 − 5 t + 6 ) y ″ − ( 2 t − 3 ) y ′ − 8 y = 0 or ( t 2 − 5 t + 6 ) y ″ − ( t − 1.5 ) y ′ − 4 y = 0 Now for the value of x factorized the coefficient of y ″ . 2 ( t 2 − 5 t + 6 ) = 0 t 2 − 2 t − 3 t + 6 = 0 t ( t − 2 ) − 3 ( t − 2 ) = 0 Further simplification of above equation. ( t − 2 ) ( t − 3 ) = 0 t 1 = 2 t 2 = 3 Substitute the value of t 1 and t 2 in x = t − t 1 t 2 − t 1 . x = t − 2 3 − 2 = t − 2 1 = t − 2 When compare the given second order equation 2 ( t 2 − 5 t + 6 ) y ″ − ( 2 t − 3 ) y ′ − 8 y = 0 with ( t 2 + A t + B ) y ″ + ( C t + D ) y ′ + K y = 0 then the value C = 1 , D = − 1.5 and K = − 4 . Now substitute the value C = 1 in equation C = a + b + 1 . C = a + b + 1 a + b + 1 = 1 a + b = 0 a = − ( b ) Substitute, the value a = − ( b ) in a b = − 4 . ( − b ) ( b ) = − 4 b 2 = 4 b = 2 Now substitute the value b = 2 in a = − b . a = − 2 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 ) . 1 × 2 − 1.5 = − c ( t 2 − t 1 ) c ( 3 − 2 ) = − 0.5 c = − 0.5 Substitute the all coefficient value in general ordinary differential equation x ( x − 1 ) y ″ + [ c − ( a + b + 1 ) x ] y ′ − a b y = 0 . x ( x − 1 ) y ″ + [ − 0.5 − ( − 2 + 2 + 1 ) x ] y ′ + 4 y = 0 x ( x − 1 ) y ″ + [ − 0.5 − ( 0 + 1 ) x ] y ′ + 4 y = 0 x ( x − 1 ) y ″ − ( 0.5 + x ) y ′ + 4 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 ( a , b , c ; x ) + A 2 x 1 − c F ( a − c + 1 , b − c + 1 , 2 − c ; x ) = A 1 F ( − 2 , 2 , − 0.5 ; t − 2 ) + A 2 x 1 + 0.5 F ( − 2 + 0.5 + 1 , 2 + 0.5 + 1 , 2 + 0.5 ; t − 2 ) = A 1 F ( − 2 , 2 , − 0.5 ; t − 2 ) + A 2 x 1.5 F ( − 0.5 , 3.5 , 2.5 ; t − 2 ) Therefore, the solution of the ordinary differential equation is y = A 1 F ( − 2 , 2 , − 0.5 ; t − 2 ) + A 2 x 1.5 F ( − 0.5 , 3.5 , 2.5 ; t − 2 ) _ .

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3. Consider the following theorem: Theorem: If n is an odd integer, then n³ is an odd integer. Note: There is an implicit universal quantifier for this theorem. Technically we could write: For all integers n, if n is an odd integer, then n³ is an odd integer. (a) Explore the statement by constructing at least three examples that satisfy the hypothesis, one of which uses a negative value. Verify the conclusion is true for each example. You do not need to write your examples formally, but your work should be easy to follow. (b) Pick one of your examples from part (a) and complete the following sentence frame: One example that verifies the theorem is when n = We see the hypothesis is true because and the conclusion is true because (c) Use the definition of odd to construct a know-show table that outlines the proof of the theorem. You do not need to write a proof at this time.

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