Explanation Given: The given initial value problem is y ″ + x y ′ + 2 y = 0 , y ( 0 ) = 3 , y ′ ( 0 ) = − 2 . Theorem used: Frobeniu’s Theorem: For a differential equation a 2 ( x ) y ″ + a 1 ( x ) y ′ + a 0 ( x ) y = 0 , there exists at least one solution of the form as follows: y = ( x − x 0 ) r ∑ n = 0 ∞ c n ( x − x 0 ) n = ∑ n = 0 ∞ c n ( x − x 0 ) n + r Calculation: The series solution for a differential equation is given as follows: y = ∑ n = 0 ∞ c n x n . Differentiate the above equation. y ′ = ∑ n = 1 ∞ n c n x n − 1 Again differentiate the above equation. y ″ = ∑ n = 2 ∞ n ( n − 1 ) c n x n − 2 Substitute the values of y , y ′ and y ″ in y ″ + x y ′ + 2 y = 0 . y ″ + x y ′ + 2 y = ∑ n = 2 ∞ n ( n − 1 ) c n x n − 2 ︸ k = n − 2 + ∑ n = 1 ∞ n c n x n ︸ k = n + 2 ∑ n = 0 ∞ c n x n ︸ k = n = ∑ k = 0 ∞ ( k + 2 ) ( k + 1 ) c k + 2 x k + ∑ k = 1 ∞ k c k x k + 2 ∑ k = 0 ∞ c k x n Further solve the above expression. y ″ + x y ′ + 2 y = 2 c 2 + 2 c 0 + ∑ k = 1 ∞ [ ( k + 2 ) ( k + 1 ) c k + 2 + ( k + 2 ) c k ] x k = 0 For the above equation to be zero, it has to satisfy two equations as shown below. 2 c 2 + 2 c 0 = 0 and ( k + 2 ) ( k + 1 ) c k + 2 + ( k + 2 ) c k = 0 The equation 2 c 2 + 2 c 0 = 0 implies that, 2 c 2 + 2 c 0 = 0 c 2 = − c 0 And the equation ( k + 2 ) ( k + 1 ) c k + 2 + ( k + 2 ) c k = 0 implies that, ( k + 2 ) ( k + 1 ) c k + 2 = − ( k + 2 ) c k c k + 2 = − 1 k + 1 c k for k = 1 , 2 , 3........ Choose c 0 = 1 and c 1 = 0 . Then, c 2 = − c 0 = − 1 and c 3 = c 1 + 2 = − 1 1 + 1 c 1 = 0 and similarly c 3 = c 5 = c 7 =

A First Course in Differential Equations with Modeling Applications (MindTap Course List)

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The solution of the initial value problem y″+xy′+2y=0 , y(0)=3 , y′(0)=−2.

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The solution of the initial value problem y ″ + x y ′ + 2 y = 0 , y ( 0 ) = 3 , y ′ ( 0 ) = − 2 .

Solution Summary: The author explains Frobeniu's Theorem: For a differential equation a_2(x)y

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The solution of the initial value problem y″+xy′+2y=0 , y(0)=3 , y′(0)=−2.

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