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Question Number 169364 by Mastermind last updated on 29/Apr/22

S h o w t h a t t h e d i f f e r e n t i a l e q u a t i o n

y^{″} - 4 y^{'} + 4 y = 0 i s s a t i s f i e d w h e n

y = {x e}^{2 x}

M a s t e r m i n d

Answered by Mathspace last updated on 29/Apr/22

(c e) \to r^{2} - 4 r + 1 = 0 \Rightarrow {(r - 2)}^{2} = 0 \Rightarrow r = 2

a n d 2 i s d o u b l e r o o t \Rightarrow y = (λ x + α) e^{2 x}