To determine The solution of the given IVP. Explanation Given: The given IVP is ( D 3 − 2 D 2 − 3 D ) y = 74 e − 3 x sin x , y ( 0 ) = − 1.4 , y ′ ( 0 ) = 3.2 , y ″ ( 0 ) = − 5.2 . Procedure used: Let y ( n ) + a n − 1 ( x ) y ( n − 1 ) + ⋯ + a 1 ( x ) y ′ + a 0 ( x ) y = r ( x ) be the n th order homogeneous ODE with constant coefficients, where y ( n ) = d n y d x n is the first term and r ( x ) ≠ 0 . A general solution of the above equation on an open interval I of the x -axis is of the form r ( x ) ≠ 0 Obtain the characteristic equation by substituting y = e λ x in the ODE as follows. y ( x ) = y h ( x ) + y p ( x ) Here y h ( x ) is a general solution of the corresponding homogeneous ODE. And y p ( x ) is any solution of the given equation on I containing no arbitrary constants. Calculation: Consider the given IVP, ( D 3 − 2 D 2 − 3 D ) y = 74 e − 3 x sin x , y ( 0 ) = − 1.4 , y ′ ( 0 ) = 3.2 , y ″ ( 0 ) = − 5.2 y ′ ′ ′ − 2 y ″ − 3 y ′ = 74 e − 3 x sin x , y ( 0 ) = − 1.4 , y ′ ( 0 ) = 3.2 , y ″ ( 0 ) = − 5.2 Consider the corresponding homogeneous equation of the given ODE, y ′ ′ ′ − 2 y ″ − 3 y ′ = 0 Obtain the characteristic equation as follows. λ 3 − 2 λ 2 − 3 λ = 0 Compute the roots of the characteristic equation as follows. λ ( λ 2 − 2 λ − 3 ) = 0 λ ( λ − 3 ) ( λ + 1 ) = 0 λ 1 = 0 , λ 2 = − 1 , λ 3 = 3 The homogeneous solution of the given ODE is given as follows. y h = c 1 e 0 x + c 2 e − x + c 3 e 3 x = c 1 + c 2 e − x + c 3 e 3 x Let the particular solution is of the form y p = e − 3 x ( A cos x + B sin x ) . Obtain the first three derivatives of the above equation as follows. y ′ p = − 3 e − 3 x ( A cos x + B sin x ) + e − 3 x ( − A sin x + B cos x ) = e − 3 x ( cos x ( B − 3 A ) + sin x ( − A − 3 B ) ) y ″ p = − 3 e − 3 x ( cos x ( B − 3 A ) + sin x ( − A − 3 B ) ) + e − 3 x ( sin x ( 3 A − B ) + cos x ( − A − 3 B ) ) = e − 3 x ( cos x ( 8 A − 6 B ) + sin x ( 6 A + 8 B ) ) y ′ ′ ′ p = − 3 e − 3 x ( cos x ( 8 A − 6 B ) + sin x ( 6 A + 8 B ) ) + e − 3 x ( sin x ( 6 B − 8 A ) + cos x ( 6 A + 8 B ) ) = e − 3 x ( cos x ( 26 B − 18 A ) + sin x ( − 26 A − 18 B ) ) Substitute the above derivatives in the given ODE as follows

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A graph G of order 12 has vertex set V(G) = {c1, c2, …, c12} for the twelve configurations inFigure 1.4. A “move” on this checkerboard corresponds to moving a single coin to anunoccupied square, where(1) the gold coin can only be moved horizontally or diagonally,(2) the silver coin can only be moved vertically or diagonally.Two vertices ci and cj (i ≠ j) are adjacent if it is possible to move ci to cj by a single move. (a) What vertices are adjacent to c1 in G?(c) Draw the subgraph of G induced by {c2, c6, c9, c11}.

i) Consider the set S = {−6, −3, 0, 3, 6}. Draw a graph G whose set of verti- ces be S and such that for i, j ∈ S, ij ∈ E(G) if ij are related to a rule that t'u you choose to apply to i and j. (ii) A graph G of order 12 has as a set of vertices c1, c2, . . . , c12 for the do- ce configurations of figure 1. A movement on said board corresponds to moving a coin to an unoccupied square using the following two rules: 1. the gold coin can move only horizontally or diagonally, 2. the silver coin can move only vertically or diagonally. Two vertices ci, cj, i̸ = j are adjacent if it is possible to move ci to cj in a single movement. a) What vertices are adjacent to c1 in G? b) Draw the subgraph induced by {c2, c6, c9, c11}

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