Given the linear differential equation y2y0 a find a...

Given the linear differential equation y′′−2y′=0: a) Find a recursion relation for a power series solution of this equation about the ordinary point x0=0. b) Solve this recursion relation to find two linearly independent solutions for this differential equation. PLEASE SHOW ALL STEPS AND ANSWERS MUST BE CORRECT FOR UPVOTE :)

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00:01 Hi, in this question we have been given an equation that is y double dash plus 3y dash is equal to 0.

00:12 Consider the condition y of 0 is equal to a 0, y dash of 0 is equal to a 1.

00:18 Let y will be equal to summation n greater than equal to 0 a n x raised 2 n, y dash will be equal to summation n greater than equal to 0 a n x raised 2 n.

00:31 N greater than equal to 0 n plus 1 a n plus 1 x raised to n and y double dash will be equal to summation n greater than equal to 0 into n plus 1 into n plus 2 a n plus 2 a n plus 2 x raised to n therefore summation n greater than equal to 0 n plus 1 into n plus 2 a n plus 2 x raise to n plus 3 submission n greater than equal to 0 n plus 1 gain plus 1 x raised to n is equal to 0 so summation n greater than equal to 0 n plus 1 into n plus 2 a n plus 2 plus 3 into n plus 1n plus 1 into x raised 2 n is equal to 0 therefore n plus 1 into n plus 2 a n plus 2 plus 3 n plus 1 a n plus 1 is equal to 0 for n greater than equal to 0 or we'll say that n minus 1 n minus 1 a n plus 3 into n minus 1 a n minus 1 is equal to 0 which implies that a n will be equal to minus 3 by n, a n minus 1.

01:54 Now we can say that for n greater than equal to 2, notice that this implies a dependency for all the an beyond n is equal to 1 on a1, while a0 takes on whatever initial value y of 0 is given.

02:11 So it can be said that in particular, a2 is equal to minus 3 divided by 2, a1, a3 will be equal to minus 3 divided by 3a2 which is equal to 3 squared divided by 3 into 2a1 that is equal to 3 square divided by 3 factorial a 1.

02:44 So we will get a4 as minus 3 cube divided by 4 factorial a 1.

02:54 Therefore, an is equal to minus 3 raise to n minus 1 divided by n factorial into a1.

03:00 The two fundamental solutions, y1, y2, is such that y is equal to y1 plus y2 where y1 is equal to a, 2, y2 is equal to minus 3 a 1, summation n greater than equal to 1 minus 3 x raised 2 n divided by n factorial.

03:16 Recall e raise 2 x is equal to summation n greater than equal to 0 x raised to n divided by n factorial...

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