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Question Number 1898 by 123456 last updated on 22/Oct/15

\frac{d f}{d t} = α f + β t + γ

f (t) = ? ?

Answered by Yozzy last updated on 22/Oct/15

\frac{d f}{d t} = α f + β t + γ w h e r e I a s s u m e t h a t α, β, γ a r e c o n s t a n t s . T h i s e q u a t i o n m a y b e

r e w r i t t e n a s \frac{d f}{d t} - α f = β t + γ (*) . T h e e q u a t i o n i s a f i r s t o r d e r l i n e a r n o n - h o m o g e n e o u s

d i f f e r e n t i a l e q u a t i o n w h i c h m a y b e s o l v e d b y u s e o f a n i n t e g r a t i n g f a c t o r ψ (t) .

L e t ψ (t) = e^{\int - α d t} = e^{- α t} . M u l t i p l y i n g b o t h s i d e s o f (*) b y ψ (t) w e g e t

ψ (t) \frac{d f}{d t} + ψ (t) (- α) f = ψ (t) (β t + γ)

\Rightarrow e^{- α t} \frac{d f}{d t} + (- α e^{- α t}) f = e^{- α t} (β t + γ)

N o t i c e t h a t \frac{d}{d t} (e^{- α t} f) = e^{- α t} \frac{d f}{d t} + (- α e^{- α t}) f .

∴ \frac{d}{d t} ({f e}^{- α t}) = e^{- α t} (β t + γ) (* *)

I n t e g r a t i n g b o t h s i d e s o f (* *) w e g e t

{f e}^{- α t} = \int e^{- α t} (β t + γ) d t

f = e^{α t} \int e^{- α t} (β t + γ) d t

B y i n t e g r a t i n g b y p a r t s,

\int e^{- α t} (β t + γ) d t = \frac{e^{- α t}}{- α} (β t + γ) - \int \frac{e^{- α t}}{- α} \times β d t (α \neq 0)

= \frac{e^{- α t} (β t + γ)}{- α} + \frac{β}{α} \times \frac{e^{- α t}}{- α} + D

\int e^{- α t} (β t + γ) d t = \frac{e^{- α t}}{- α} (β t + γ + \frac{β}{α}) + D

∴ f = e^{α t} (\frac{e^{- α t}}{- α} (β t + \frac{β}{α} + γ) + D)

f (t) = {D e}^{α t} - \frac{1}{α} \times \frac{β α t + β + α γ}{α}

f (t) = {D e}^{α t} - \frac{β α t + β + α γ}{α^{2}} (α \neq 0) w h e r e D i s a n a r b i t r a r y c o n s t a n t .

C h e c k i n g S o l u t i o n :

\frac{d f}{d t} = α {D e}^{α t} - \frac{β}{α}

{D e}^{α t} = f + \frac{β}{α} t + \frac{β + α γ}{α^{2}}

\Rightarrow \frac{d f}{d t} = α f + β t + α \frac{β}{α^{2}} + α \times \frac{α γ}{α^{2}} - \frac{β}{α}

\frac{d f}{d t} = α f + β t + γ a s g i v e n .

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