A general solution in terms of hypergeometric functions. The solution of the ordinary differential equation is y = A 1 F ( 0.33 , − 1 , 0.33 ; t + 1 ) + A 2 ( t + 1 ) 0.66 F ( − 0.33 , 1 , 0.67 ; t + 1 ) _ . 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 1 + D = − c ( t 2 − t 1 ) , C = a + b + 1 and K = a b . The value of F ( a , b , c ; x ) is equal to F ( b , a , c ; x ) , does not affect the solution of ordinary differential equation. Calculation: The given system of equation is as 3 t ( 1 + t ) y ″ + t y ′ − y = 0 or t ( 1 + t ) y ″ + 0 . 3 3 t y ′ − 0 . 3 3 y = 0 Now for the value of x factorized the coefficient of y ″ . t ( 1 + t ) = 0 t 1 = − 1 t 2 = 0 Substitute the value of t 1 and t 2 in x = t − t 1 t 2 − t 1 . x = t − ( − 1 ) 0 − ( − 1 ) = t + 1 1 = t + 1 When compare the given second order equation t ( 1 + t ) y ″ + 0 . 3 3 t y ′ − 0 . 3 3 y = 0 with ( t 2 + A t + B ) y ″ + ( C t + D ) y ′ + K y = 0 then the value C = 0.33 , D = 0 and K = − 0.33 . Now substitute the value C = 0.33 in equation C = a + b + 1 . C = a + b + 1 a + b + 1 = 0.33 a + b = − 0.66 a = − 0.66 − b Substitute, the value a = − 0.66 − b in a b = − 0.33 . ( − 0.66 − b ) b = − 0.33 − 0.66 b − b 2 = − 0.33 b 2 + 0.66 b − 0.33 = 0 Further simplification of above equation. b = − 0.66 ± 0.66 2 − 4 × 1 × ( − 0.33 ) 2 × 1 = − 0.66 ± 1.98 2 × 1 = − 0.66 ± 1.40 2 = 0.37 , − 1.03 Now substitute the value b = 0.37 in a = − 0.66 − b . a = − 0.66 − 0.37 = − 1.03 ≃ − 1 So, the value of a = − 1 , b = 0.37 or b = − 1 , a = 0.37 . Because the function F ( a , b , c ; x ) = F ( b , a , c ; x ) . 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 ) . 0.33 ( − 1 ) + 0 = − c ( 0 + 1 ) c = 0.33 Substitute the all coefficient value in general ordinary differential equation x ( 1 − x ) y ″ + [ c − ( a + b + 1 ) x ] y ′ − a b y = 0 . x ( 1 − x ) y ″ + [ 0.33 − ( − 0.66 + 1 ) x ] y ′ − ( − 0.33 ) y = 0 x ( 1 − x ) y ″ + [ 0.33 − 0.33 x ] y ′ + ( 0.33 ) 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 ( b , a , c ; x ) + A 2 x 1 − c F ( a − c + 1 , b − c + 1 , 2 − c ; x ) = A 1 F ( 0.33 , − 1 , 0.33 ; t + 1 ) + A 2 x 1 − 0.33 F ( − 1 − 0.33 + 1 , 0.33 − ( 0.33 ) + 1 , 2 − ( 0.33 ) ; t + 1 ) = A 1 F ( 0.33 , − 1 , 0.33 ; t + 1 ) + A 2 x 0.66 F ( − 0.33 , 1 , 0.67 ; t + 1 ) Therefore, the solution of the ordinary differential equation is y = A 1 F ( 0.33 , − 1 , 0.33 ; t + 1 ) + A 2 ( t + 1 ) 0.66 F ( − 0.33 , 1 , 0.67 ; t + 1 ) _ .

Question

Want to see the full answer?

Answer 1

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

7. Define the sequence {b} by bo = 0 Ել ։ = 2 8. bn=4bn-1-4bn-2 for n ≥ 2 (a) Give the first five terms of this sequence. (b) Prove: For all n = N, bn = 2nn. Let a Rsuch that a 1, and let nЄ N. We're going to derive a formula for Σoa without needing to prove it by induction. Tip: it can be helpful to use C1+C2+...+Cn notation instead of summation notation when working this out on scratch paper. (a) Take a a² and manipulate it until it is in the form Σ.a. i=0 (b) Using this, calculate the difference between a Σ0 a² and Σ0 a², simplifying away the summation notation. i=0 (c) Now that you know what (a – 1) Σ0 a² equals, divide both sides by a − 1 to derive the formula for a². (d) (Optional, just for induction practice) Prove this formula using induction.

3. Let A, B, and C be sets and let f: A B and g BC be functions. For each of the following, draw arrow diagrams that illustrate the situation, and then prove the proposition. (a) If ƒ and g are injective, then go f is injective. (b) If ƒ and g are surjective, then go f is surjective. (c) If gof is injective then f is injective. Make sure your arrow diagram shows that 9 does not need to be injective! (d) If gof is surjective then g is surjective. Make sure your arrow diagram shows that f does not need to be surjective!

4. 5. 6. Let X be a set and let f: XX be a function. We say that f is an involution if fof idx and that f is idempotent if f f = f. (a) If f is an involution, must it be invertible? Why or why not?2 (b) If f is idempotent, must it be invertible? Why or why not? (c) If f is idempotent and x E range(f), prove that f(x) = x. Prove that [log3 536] 5. You proof must be verifiable by someone who does not have access to a scientific calculator or a logarithm table (you cannot use log3 536≈ 5.7). Define the sequence {a} by a = 2-i for i≥ 1. (a) Give the first five terms of the sequence. (b) Prove that the sequence is increasing.