Solve-the-d-e-sin-d-3-y-dt-3-3t-2-y-6t-

Question Number 2160 by Yozzis last updated on 05/Nov/15

S o l v e t h e d . e

s i n (\frac{d^{3} y}{{d t}^{3}}) + 3 t^{2} y = 6 t .

Commented by Yozzi last updated on 07/Nov/15

I w a s h e l p i n g a f r i e n d w i t h h i s

e n g i n e e r i n g m a t h h o m e w o r k a n d

h e t h e r e f o r e t o o k a p i c t u r e o f t h e p r o b l e m .

I t w a s a w o r k s h e e t . I a n s w e r e d t h e q u e s t i o n b u t

b u t i n t h a t p i c t u r e I s a w a n u m b e r o f

s t r a n g e D . E s a s p a r t o f a n o t h e r e x e r c i s e . T h e e x e r c i s e w a s t o s i m p l y

i d e n t i f y t h e t y p e o f D . E b u t I t h o u g h t,

^{'} H o w w o u l d o n e s o l v e i t o r p r o v e i t i s

u n s o l v a b l e ?^{'} T h e D . E p r o b l e m a b o v e

t h i s p r o b l e m w a s a l s o f r o m t h e s e t

o f s t r a n g e D . E s I s a w .

Y o z z i s = Y o z z i

Commented by prakash jain last updated on 07/Nov/15

Difficult . Did you come across this while

working on physics problem ?

Commented by prakash jain last updated on 08/Nov/15

a s s u m i n g y c a n b e d i f f e r e n t i a t e d 4 t i m e s

\cos (\frac{d^{3} y}{{d t}^{4}}) \frac{d^{4} y}{{d t}^{4}} = 6 - 6 t y - 3 t^{2} \frac{d y}{d t}

\cos (\frac{d^{3} y}{{d t}^{4}}) = {(\frac{d^{4} y}{{d t}^{4}})}^{- 1} (6 - 6 t y - 3 t^{2} \frac{d y}{d t})

1 = 36 t^{2} + {(\frac{d^{4} y}{{d t}^{4}})}^{(- 2)} {(6 - 6 t y - 3 t^{2} \frac{d y}{d t})}^{2}

The above equation appears solvable with

series expansion .

Answered by prakash jain last updated on 08/Nov/15

Trying with variable substitution

to remove \sin function of a differnetial .

y = \frac{2}{t} - \frac{1}{3 t^{2}} \sin u

\sin (\frac{d^{3} y}{{d t}^{3}}) = 6 t - 3 t^{2} y = 6 t - 6 t + \sin u

\frac{d^{3} y}{{d t}^{3}} = u

c o m p u t i n g \frac{d^{3} y}{{d t}^{3}} i n t e r m s o f u a n d t

\frac{d y}{d t} = - \frac{2}{t^{2}} + \frac{2}{3 t^{3}} \sin u - \frac{1}{3 t^{2}} \cos u \frac{d u}{d t}

\frac{d^{2} y}{{d t}^{2}} = \frac{4}{t^{3}} + \frac{2}{3 t^{3}} \cos u \frac{d u}{d t} - \frac{2}{t^{4}} \sin u + \frac{2}{3 t^{3}} \cos u \frac{d u}{d t}

- \frac{1}{3 t^{2}} \cos u \frac{d^{2} u}{{d t}^{2}} + \frac{1}{3 t^{2}} \sin u {(\frac{d u}{d t})}^{2}

= \frac{4}{t^{3}} + \frac{4}{3 t^{3}} \cos u \frac{d u}{d t} - \frac{2}{t^{4}} \sin u - \frac{1}{3 t^{2}} \cos u \frac{d^{2} u}{{d t}^{2}} + \frac{1}{3 t^{2}} \sin u {(\frac{d u}{d t})}^{2}

\frac{d^{3} y}{{d t}^{3}} = - \frac{12}{t^{4}} - \frac{4}{t^{4}} \cos u \frac{d u}{d t} + \frac{4}{3 t^{3}} \cos u \frac{d^{2} u}{{d t}^{2}} - \frac{4}{3 t^{3}} \sin u {(\frac{d u}{d t})}^{2}

+ \frac{8}{t^{5}} \sin u - \frac{2}{t^{4}} \cos u \frac{d u}{d t}

+ \frac{2}{3 t^{3}} \cos u \frac{d^{2} u}{{d t}^{3}} + \frac{1}{3 t^{2}} \sin u (\frac{d u}{d t}) - \frac{1}{3 t^{2}} \cos u \frac{d^{3} u}{{d t}^{3}}

+ \frac{4}{t^{4}} \sin u {(\frac{d u}{d t})}^{2} - \frac{4}{3 t^{3}} \cos u {(\frac{d u}{d t})}^{3} - \frac{4}{3 t^{3}} \sin u (2 \frac{d u}{d t}) (\frac{d^{2} u}{{d t}^{2}})

w e g e t a n e q u a t i o n b e t w e e n t a n d u .

s t i l l a v e r y d i f f i c u l t e q u a t i o n