D-2-1-2-y-t-3-find-solution-

Question Number 82390 by jagoll last updated on 21/Feb/20

{(D^{2} - 1)}^{2} y = t^{3}

f i n d s o l u t i o n

Answered by TANMAY PANACEA last updated on 21/Feb/20

y = e^{m t} s o \frac{d y}{d t} = {m e}^{m t} a n d \frac{d^{2} y}{{d t}^{2}} = m^{2} e^{m t}

{(m^{2} - 1)}^{2} = 0

m = 1, 1, - 1, - 1

c o m p l e m e n t a r y f u n c t i o n

y = {A e}^{t} + {B t e}^{t} + {C e}^{- t} + {F t e}^{- t}

p a r t i c u l a r i n t r e g a l

n o w f o r m u l a

{(1 + x)}^{- 2} = 1 - 2 x + 3 x^{2} - 4 x^{3} + \dots

y = \frac{t^{3}}{{(D^{2} - 1)}^{2}} = \frac{t^{3}}{{(1 - D^{2})}^{2}}

y = {(1 - D^{2})}^{- 2} t^{3}

= (1 + 2 D^{2} + 3 D^{4} + o t h e r s t e r m s i g n o r e d) t^{3}

= (t^{3} + 2 \times 3 \times 2 t) [\frac{{d t}^{3}}{d t} = 3 t^{2} \to \frac{d (3 t^{2})}{d t} = 6 t]

= t^{3} + 12 t

c o m p l e t e s o l u t i o n

y = {A e}^{t} + {B t e}^{t} + {C e}^{- t} + {F t e}^{- t} + t^{3} + 12 t

Commented by jagoll last updated on 21/Feb/20

t h a n k y o u s i r

Commented by TANMAY PANACEA last updated on 21/Feb/20

m o s t w e l c o m e