Question-97891

Question Number 97891 by pranesh last updated on 10/Jun/20

Commented by mr W last updated on 10/Jun/20

y = e^{x} - 1

Commented by bemath last updated on 10/Jun/20

λ^{2} - 1 = 0

λ = \pm 1

y_{h} = {Ae}^{x} + {Be}^{- x}

particular y_{p} = a x + b

y^{'} = a, y^{″} = 0

0 - (a x + b) = 1 \Leftrightarrow {\begin{cases} a = 0 \\ b = - 1 \end{cases}

y_{c} = A e^{x} + {Be}^{- x} - 1

x = 0, y = 0

\Leftrightarrow 0 = A + B - 1, A + B = 1

y_{c} = {Ae}^{x} + (1 - A) e^{- x} - 1

Commented by mr W last updated on 10/Jun/20

A m u s t b e e q u a l t o 1, o t h e r w e i s e

\lim_{x \to - \infty} y = 0 + (1 - A) \infty - 1 = \infty \neq a

Commented by bemath last updated on 10/Jun/20

how to get A = 1 ?

Commented by mr W last updated on 10/Jun/20

s u c h t h a t \lim_{x \to - \infty} y ↛ \infty, t h e t e r m

e^{- x} m u s t v a n i s h, t h e r e f o r e 1 - A = 0 .

Commented by bemath last updated on 10/Jun/20

oo yes sir . thank you

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