The first four non-zero terms in a power series expansion about x = 0 for a general solution to the given differential equation. The required first four non-zero terms in a power series expansion about x = 0 are a 0 , a 1 x , a 0 6 x 2 , a 0 27 x 3 . Given: 2 x − 3 y ' ' − x y ' + y = 0 Concept used: For power series y ( x ) = ∑ n = 0 ∞ a n x n ... i , Its first derivative : y ' x = ∑ n = 1 ∞ n a n x n − 1 and second derivative: y ' ' x = ∑ n = 2 ∞ n n − 1 a n x n − 2 … i i Calculation: Substitute i and i i in the given equation. 2 x − 3 ∑ n = 2 ∞ n n − 1 a n x n − 2 − x ∑ n = 1 ∞ n a n x n − 1 + ∑ n = 0 ∞ a n x n = 0 2 x ∑ n = 2 ∞ n n − 1 a n x n − 2 − 3 ∑ n = 2 ∞ n n − 1 a n x n − 2 − ∑ n = 1 ∞ n a n x n − 1 + 1 + ∑ n = 0 ∞ a n x n = 0 ∑ n = 2 ∞ 2 n n − 1 a n x n − 1 − 3 ∑ n = 2 ∞ n n − 1 a n x n − 2 − ∑ n = 1 ∞ n a n x n + ∑ n = 0 ∞ a n x n = 0 Assume k = n − 1 in the first term and k = n − 2 in the second term. ∑ k = 1 ∞ 2 k k + 1 a k + 1 x k − 3 ∑ k = 0 ∞ k + 1 k + 2 a k + 2 x k − ∑ k = 1 ∞ k a k x k + ∑ k = 0 ∞ a k x k = 0 ∑ k = 1 ∞ 2 k k + 1 a k + 1 x k − 3 0 + 1 0 + 2 x 0 a 0 + 2 + 3 ∑ k = 1 ∞ k + 1 k + 2 a k + 2 x k − ∑ k = 1 ∞ k a k x k + a 0 x 0 + ∑ k = 1 ∞ a k x k = 0 ∑ k = 1 ∞ 2 k k + 1 a k + 1 x k − 6 a 2 + 3 ∑ k = 1 ∞ k + 1 k + 2 a k + 2 x k − ∑ k = 1 ∞ k a k x k + a 0 + ∑ k = 1 ∞ a k x k = 0 a 0 − 6 a 2 + ∑ k = 1 ∞ 2 k k + 1 a k + 1 x k − 3 ∑ k = 1 ∞ k + 1 k + 2 a k + 2 x k − ∑ k = 1 ∞ k a k x k + ∑ k = 1 ∞ a k x k = 0 a 0 − 6 a 2 + ∑ k = 1 ∞ 2 k k + 1 a k + 1 − 3 k + 1 k + 2 a k + 2 − k a k + a k x k = 0 Equate all coefficients to zero and simplify. a 0 − 6 a 2 = 0 a 0 = 6 a 2 a 2 = a 0 6 2 k k + 1 a k + 1 − 3 k + 1 k + 2 a k + 2 − k a k + a k = 0 , where k ≥ 1 3 k + 1 k + 2 a k + 2 = 2 k k + 1 a k + 1 − k a k + a k , where k ≥ 1 a k + 2 = 2 k k + 1 a k + 1 − k a k + a k 3 k + 1 k + 2 , where k ≥ 1 For k = 1 , a 3 = 2 2 a 2 − a 1 + a 1 3 2 3 = 4 a 2 18 = 2 a 9 = 2 a 0 6 9 a 2 = a 0 6 = a 0 27 For k = 2 , a 4 = 2 2 3 a 3 − 2 a 2 + a 2 3 3 4 = 12 a 3 − a 2 36 = a 0 27 3 − a 0 6 36 a 3 = a 0 27 , a 2 = a 0 6 = a 0 81 − a 0 216 = 5 a 0 684 Thus, y x = a 0 + a 1 x + a 2 x 2 + a 3 x 3 + ... = a 0 + a 1 x + a 0 6 x 2 + a 0 27 x 3 + 5 a 0 648 x 4 + ... Therefore, the required first four non-zero terms in a power series expansion about x = 0 are a 0 , a 1 x , a 0 6 x 2 , a 0 27 x 3 .

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Q1.4 1 Point V=C(R), the vector space of all real-valued continuous functions whose domain is the set R of all real numbers, and H is the subset of C(R) consisting of all of the constant functions. (e.g. the function ƒ : R → R defined by the formula f(x) = 3 for all x E R is an example of one element of H.) OH is a subspace of V. H is not a subspace of V. Save Answer

Example 3.2. Solve the following boundary value problem by ADM (Adomian decomposition) method with the boundary conditions მი მი z- = 2x²+3 дг Əz w(x, 0) = x² - 3x, θω (x, 0) = i(2x+3). ay

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