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When solving a fixed-constant linear ordinary differential equation where the characteristic equation has repeated roots, why do we get the next independent solution in the form of x n e mx ? Show this through an example. Let’s suppose we want to solve the ordinary differential equation d2y dy + 6 + 9y = 0 2 dx dx The characteristic equation is m 2 + 6m1 + 9m 0 = 0 m 2 + 6m + 9 = 0 (m + 3) 2 = 0 The solution to the characteristic equation is m = −3,−3 and the homogeneous part of the solution to the ordinary differential equation is y H = K 1e −3 x + K 2 xe −3 x Since the forcing function is zero, the particular part of the solution to the ordinary differential equation is yP = 0 The complete solution to the ordinary differential equation is y = yH + yP = K1e −3 x + K 2 xe −3 x But why is K 2 xe −3 x part of the solution? See the answer below. d2y dy + 6 + 9y = 0 2 dx dx can be rewritten as d d + 3 + 3 y = 0 dx dx (1) d + 3 y = z dx (2) Let Then equation (1) can be written as d + 3 z = 0 dx (3) Let z = K 2 e m2 x be a possible solution form of equation (3), then d m x + 3 K 2 e 2 = 0 dx ( ) K 2 m 2 e m2 x + 3 K 2 e m2 x = 0 K 2 (m2 + 3)e m2 x = 0 Then m2 = −3 is a solution which is nontrivial. So z = K 2 e −3 x From equation (2) d + 3 y = z dx d −3 x + 3 y = K 2 e dx dy + 3 y = K 2 e −3 x dx Multiply both sides by e 3 x dy e 3 x + 3 y = K 2 e −3 x e 3 x dx e3x dy + 3e 3 x y = K 2 dx e3x d d ( y ) + y (e 3 x ) = K 2 dx dx By chain rule d 3x e y = K2 dx ( ) Integrating both sides with respect to x , we get e 3 x y = K 2 x + K1 y = K 2 xe −3 x + K 1e −3 x