Explanation Given: The equation is, ( 1 − x 2 ) y ″ − 2 x y ′ + 2 y = 0 ... ( 1 ) Approach: The linear second-order differential equation given by, ( 1 − x 2 ) y ″ − 2 x y ′ + n ( n + 1 ) y = 0 , n is a fixed parameter is called Legendre’s Equation. Set z = x − 1 , the equation is transformed as, z ( z + 2 ) d 2 y d z 2 + 2 ( z + 1 ) d y d z − n ( n + 1 ) y = 0 ... ( 2 ) The indicial equation for equation ( 2 ) is given as, r ( r − 1 ) + r = 0 The solution is given as, y 1 ( x ) = 1 + ∑ k = 1 ∞ ( − n ) k ( n + 1 ) k k ! ( 1 ) k ( 1 − x 2 ) k Calculation: The given equation is Legendre’s equation. Let z = x − 1 Therefore, the equation ( 1 ) is transformed as, z ( z + 2 ) d 2 y d z 2 + 2 ( z + 1 ) d y d z − 2 y = 0 … ( 3 ) The indicial equation is given by, r ( r − 1 ) + r = 0 r 2 = 0 r = 0 , 0 Let y ( z ) = ∑ k = 0 ∞ a k z k be the solution to equation ( 3 ) Differentiate y ( z ) as, y ′ ( z ) = ∑ k = 0 ∞ k a k z k − 1 Again differentiate y ′ ( z ) as, y ″ ( z ) = ∑ k = 0 ∞ k ( k − 1 ) a k z k − 2 Substitute ∑ k = 0 ∞ a k z k for y ( z ) , ∑ k = 0 ∞ k a k z k − 1 for y ′ ( z ) , and ∑ k = 0 ∞ k ( k − 1 ) a k z k − 2 for y ″ ( z ) in equation ( 3 ) as, z ( z + 2 ) d 2 y d z 2 + 2 ( z + 1 ) d y d z − 2 y = 0 z ( z + 2 ) ( ∑ k = 0 ∞ k ( k − 1 ) a k z k − 2 ) + 2 ( z + 1 ) ( ∑ k = 0 ∞ k a k z k − 1 ) − 2 ( ∑ k = 0 ∞ a k z k ) = 0 Solve further to get the solution as, y 1 ( x ) = 1 − ∑ k = 1 ∞ x 2 k 2 k − 1 Let y ( x ) = v ( x ) y 1 ( x ) Substitute the expression for y 1 ( x ) as, y ( x ) = v ( x ) ( 1 − ∑ k = 1 ∞ x 2 k 2 k − 1 ) Differentiate y ( x ) with respect to x . y ′ ( x ) = v ′ ( x ) ( 1 − ∑ k = 1 ∞ x 2 k 2 k − 1 ) − v ( x ) ∑ k = 1 ∞ 2 k x 2 k − 1 2 k − 1 Again differentiate y ′ ( x ) with respect to x

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Q1) Classify the following statements as a true or false statements a. Any ring with identity is a finitely generated right R module.- b. An ideal 22 is small ideal in Z c. A nontrivial direct summand of a module cannot be large or small submodule d. The sum of a finite family of small submodules of a module M is small in M A module M 0 is called directly indecomposable if and only if 0 and M are the only direct summands of M f. A monomorphism a: M-N is said to split if and only if Ker(a) is a direct- summand in M & Z₂ contains no minimal submodules h. Qz is a finitely generated module i. Every divisible Z-module is injective j. Every free module is a projective module Q4) Give an example and explain your claim in each case a) A module M which has two composition senes 7 b) A free subset of a modale c) A free module 24 d) A module contains a direct summand submodule 7, e) A short exact sequence of modules 74.

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The general solution about x = 0 for the given equation

Solution Summary: The author explains the general solution about x=0 for the given equation. The linear second-order differential equation is called Legendre’s Equation.

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The general solution about x = 0 for the given equation

Solution Summary: The author explains the general solution about x=0 for the given equation. The linear second-order differential equation is called Legendre’s Equation.

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To find: The general solution about x=0 for the given equation

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Chapter 8 Solutions

To find:

The general solution about x=0 for the given equation