a-n-d-n-dt-n-a-n-1-d-n-1-dt-n-1-a-1-d-dt-a-0-0-Is-it-solvable-

Question Number 114722 by Dwaipayan Shikari last updated on 20/Sep/20

a_{n} \frac{d^{n} Ψ}{{d t}^{n}} + a_{n - 1} \frac{d^{n - 1} Ψ}{{d t}^{n - 1}} + \dots . . + a_{1} \frac{d Ψ}{d t} + a_{0} Ψ = 0

I s i t s o l v a b l e ? ? ?

Answered by Olaf last updated on 21/Sep/20

Only in special cases .

Not in the general case

because you need to know

the roots of the polynomial

\underset{k = 0}{\sum^{n}} a_{k} x^{k}

Answered by aleks041103 last updated on 21/Sep/20

N o t i n g e n e r a l \dots a t l e a s t n o t a n a l y t i c a l y :

l e t ψ = e^{r t} b e a s o l u t i o n

t h e n

\underset{i = 0}{\sum^{n}} a_{i} \frac{d^{i} ψ}{{d t}^{i}} = 0

\underset{i = 0}{\sum^{n}} a_{i} r^{i} e^{r t} = (\underset{i = 0}{\sum^{n}} a_{i} r^{i}) e^{r t} = 0

s i n c e e^{r t} \neq 0, t h e n

\underset{i = 0}{\sum^{n}} a_{i} r^{i} = 0

t h i s i s a p o l y n o m i a l o f d e g r e e n a n d

b y t h e f u n d a m e n t a l t h e o r y o f a l g e b r a

i t h a s n c o m p l e x s o l u t i o n s w h i c h w e

w i l l d e n o t e w i t h r_{1}, r_{2}, \dots r_{n} .

B e c a u s e t h e O D E i s l i n e a r, t h e n a n y

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

s o l u t i o n s w i l l b e a s o l u t i o n :

Ψ = \underset{k = 1}{\sum^{n}} A_{k} e^{r_{k} t}

I n g e n e r a l a n O D E o f d e g r e e n, t h a t

i s i t h a s a n n - t h d e r i v a t i v e, w i l l,

f i g i r a t i v e l y s a i d, n e e d t o b e i n t e g r a t e d

n t i m e s, w h i c h m e a n s t h a t i n t h e e n d

w e w i l l e n d u p w i t h n c o n s t a n t s o f

i n t e g r a t i o n

S i n c e o u r s o l u t i o n a l s o h a s n c o n s t a n t s

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

T o r e c a p :

\underset{i = 0}{\sum^{n}} a_{i} \frac{d^{i} Ψ}{{d t}^{i}} = 0 \Rightarrow Ψ = \underset{k = 1}{\sum^{n}} A_{k} e^{r_{k} t},

w h e r e f o r \forall k = 1, \dots n, \underset{i = 0}{\sum^{n}} a_{i} r_{k}^{i} = 0

a n d t h e c o n s t a n t s A_{k} a r e f i x e d u s i n g

t h e i n i t i a l (o r p o s s i b l y b o n d a r y)

c o n d i t i o n s .