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Question Number 125991 by bramlexs22 last updated on 16/Dec/20

I f a a n d b a r b i t a r y c o n s t a n t s,

f i n d a s e c o n d - o r d e r e q u a t i o n

w h i c h h a s y = {a e}^{x} + b \cos x a s a

g e n e r a l s o l u t i o n .

Answered by liberty last updated on 16/Dec/20

b y d i f f e r e n t i a t i n g t h e g i v e n e x p r e s s i o n

w e f i n d \to y^{'} = {a e}^{x} - b \sin x (2)

y^{″} = {a e}^{x} - b \cos x (3)

t h e n b y a d d i n g a n d s u b t r a c t i n g e q (1) a n d (3)

g i v e s {\begin{cases} a = \frac{y^{″} + y}{2 e^{x}} \\ b = \frac{y - y^{″}}{2 \cos x} \end{cases}

s u b s t i t u t i o n o f t h e s e i n t o e q (2) w e g e t

y^{'} = \frac{y + y^{″}}{2 e^{x}} . e^{x} - \frac{y - y^{″}}{2 \cos x} . \sin x

o r (1 + \tan x) y^{″} - 2 y^{'} + (1 - \tan x) y = 0 .

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