Find a solution of the form ∅(x)=x^r Σ_(k=0) ^α c_k x^k (x


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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

$\emptyset (x) = \underset{k = 0}{\sum^{α}} c_{k} x^{k + r} = c_{0} x^{r} + c_{1} x^{r + 1} + c_{2} x^{r + 2} + . . . + c_{k} x^{k + r} + . . . + c_{α} x^{α + r}$ $\emptyset^{'} (x) = \underset{k = 0}{\sum^{α}} c_{k} (k + r) x^{k + r - 1}$ $\emptyset^{'} (x) = c_{0} {r x}^{r - 1} + c_{1} (r + 1) x^{r} + c_{2} (r + 2) x^{r + 1} + c_{3} (r + 3) x^{r + 2} + . . . + c_{k} (r + k) x^{k + r - 1} + . . . + c_{α} (r + α) x^{α + r - 1}$ $\emptyset^{'} (x) - \emptyset (x) = c_{0} {r x}^{r - 1} + (c_{1} (r + 1) - c_{0}) x^{r} + (c_{2} (r + 2) - c_{1}) x^{r + 1} + . . . + (c_{k} (r + k) - c_{k - 1}) x^{r + k - 1} + . . . + (c_{α} (r + α) - c_{α - 1}) x^{r + α - 1} - c_{α} x^{r + α}$ $\emptyset^{'} (x) - \emptyset (x) = c_{0} {r x}^{r - 1} - c_{α} x^{r + α} + \underset{k = 1}{\sum^{α}} (c_{k} (r + k) - c_{k - 1}) x^{r + k - 1}$ $x \emptyset^{″} (x) = \underset{k = 0}{\sum^{α}} c_{k} (k + r) (k + r - 1) x^{k + r - 1}$ $x \emptyset^{″} (x) = c_{0} r (r - 1) x^{r - 1} + c_{1} (r + 1) {r x}^{r} + c_{2} (r + 2) (r + 1) x^{r + 1} + c_{3} (r + 3) (r + 2) x^{r + 2} + . . . + c_{k} (k + r) (k + r - 1) x^{k + r - 1} + . . . + c_{α} (r + α) (α + r - 1) x^{α + r - 1}$ $x \emptyset^{″} (x) = c_{0} r (r - 1) x^{r - 1} + \underset{k = 1}{\sum^{α}} c_{k} (k + r) (k + r - 1) x^{k + r - 1}$ $x \emptyset^{″} (x) + \emptyset^{'} (x) - \emptyset (x) = 0$ $∴ x^{r - 1} (c_{0} r^{2} - c_{0} r + c_{0} r) - c_{α} x^{r + α} + \underset{k = 1}{\sum^{α}} x^{k + r - 1} (c_{k} {(k + r)}^{2} - c_{k - 1}) = 0$ $c_{0} r^{2} x^{r - 1} - c_{α} x^{r + α} + \underset{k = 1}{\sum^{α}} x^{k + r - 1} (c_{k} (k + r) - c_{k - 1}) = 0$ $\Rightarrow c_{0} = c_{α} = 0$ $∴ c_{1} (1 + r) - c_{0} = 0 \Rightarrow c_{1} = 0$ $c_{2} (2 + r) - c_{1} = 0 \Rightarrow c_{2} = 0$ $c_{3} (3 + r) - c_{2} = 0 \Rightarrow c_{3} = 0$ $∴ c_{k} = 0 \forall k \in Z^{⩾}$ $∴ \emptyset (x) = 0 + 0 + 0 + 0 + . . . + 0 = 0$
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

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