Find-a-solution-of-the-form-x-x-r-k-0-c-k-x-k-x-gt-0-for-xy-y-y-0-

Question Number 6475 by WAI LIN last updated on 28/Jun/16

F i n d a s o l u t i o n o f t h e f o r m \emptyset (x) = x^{r} \underset{k = 0}{\sum^{α}} c_{k} x^{k} (x > 0) f o r

{x y}^{^{″}} + y^{'} - y = 0 .

Commented by Yozzii last updated on 29/Jun/16

\emptyset (x) = \underset{k = 0}{\sum^{α}} c_{k} x^{k + r} = c_{0} x^{r} + c_{1} x^{r + 1} + c_{2} x^{r + 2} + \dots + c_{k} x^{k + r} + \dots + c_{α} x^{α + r}

\emptyset^{'} (x) = \underset{k = 0}{\sum^{α}} c_{k} (k + r) x^{k + r - 1}

\emptyset^{'} (x) = c_{0} {r x}^{r - 1} + c_{1} (r + 1) x^{r} + c_{2} (r + 2) x^{r + 1} + c_{3} (r + 3) x^{r + 2} + \dots + c_{k} (r + k) x^{k + r - 1} + \dots + c_{α} (r + α) x^{α + r - 1}

\emptyset^{'} (x) - \emptyset (x) = c_{0} {r x}^{r - 1} + (c_{1} (r + 1) - c_{0}) x^{r} + (c_{2} (r + 2) - c_{1}) x^{r + 1} + \dots + (c_{k} (r + k) - c_{k - 1}) x^{r + k - 1} + \dots + (c_{α} (r + α) - c_{α - 1}) x^{r + α - 1} - c_{α} x^{r + α}

\emptyset^{'} (x) - \emptyset (x) = c_{0} {r x}^{r - 1} - c_{α} x^{r + α} + \underset{k = 1}{\sum^{α}} (c_{k} (r + k) - c_{k - 1}) x^{r + k - 1}

x \emptyset^{″} (x) = \underset{k = 0}{\sum^{α}} c_{k} (k + r) (k + r - 1) x^{k + r - 1}

x \emptyset^{″} (x) = c_{0} r (r - 1) x^{r - 1} + c_{1} (r + 1) {r x}^{r} + c_{2} (r + 2) (r + 1) x^{r + 1} + c_{3} (r + 3) (r + 2) x^{r + 2} + \dots + c_{k} (k + r) (k + r - 1) x^{k + r - 1} + \dots + c_{α} (r + α) (α + r - 1) x^{α + r - 1}

x \emptyset^{″} (x) = c_{0} r (r - 1) x^{r - 1} + \underset{k = 1}{\sum^{α}} c_{k} (k + r) (k + r - 1) x^{k + r - 1}

x \emptyset^{″} (x) + \emptyset^{'} (x) - \emptyset (x) = 0

∴ x^{r - 1} (c_{0} r^{2} - c_{0} r + c_{0} r) - c_{α} x^{r + α} + \underset{k = 1}{\sum^{α}} x^{k + r - 1} (c_{k} {(k + r)}^{2} - c_{k - 1}) = 0

c_{0} r^{2} x^{r - 1} - c_{α} x^{r + α} + \underset{k = 1}{\sum^{α}} x^{k + r - 1} (c_{k} (k + r) - c_{k - 1}) = 0

\Rightarrow c_{0} = c_{α} = 0

∴ c_{1} (1 + r) - c_{0} = 0 \Rightarrow c_{1} = 0

c_{2} (2 + r) - c_{1} = 0 \Rightarrow c_{2} = 0

c_{3} (3 + r) - c_{2} = 0 \Rightarrow c_{3} = 0

∴ c_{k} = 0 \forall k \in Z^{⩾}

∴ \emptyset (x) = 0 + 0 + 0 + 0 + \dots + 0 = 0

Commented by nburiburu last updated on 29/Jun/16

{x y}^{″} + y^{'} = {({x y}^{'})}^{'} = y

i f y = x^{w} \Rightarrow {(w)}^{2} . x^{w - 1} = x^{w} w h i c h i s i m p o s s i b l e f o r w \in R

s o t h e o n l y w a y t o c o n t i n u e w o u l d b e

t o a s s u m e w = w (x)

y = x^{w (x)} \Rightarrow {(w . x^{w})}^{'} = x^{w} [w^{'} (1 + l n x) + w^{2} / x] = x^{w}

w^{'} (1 + l n x) + \frac{w^{2}}{x} = 1

w^{'} (x + x l n x) + w^{2} = x

h o m o g e n e u s s o l u t i o n :

w^{'} (x + x l n x) + w^{2} = 0

\frac{1}{w_{h}^{2}} {d w}_{h} = \frac{- 1}{x (1 + l n x)} d x

- \frac{1}{w_{h}} = - l n (1 + l n x)

w_{h} = \frac{1}{l n (1 + l n x)}

N o t e : f o r t h e e q t o h a v e t h e s t r u c t u r e

m e n t i o n a s ϕ (x) t h e e q s h o u l d b e

x^{2} y^{″} + {x y}^{'} - y = 0

Commented by WAI LIN last updated on 01/Jul/16

U s i n g t h e p o w e r s e r i e s m e t h o d (F r o b e n i u s m e t h o d)

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