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Question Number 6475 by WAI LIN last updated on 28/Jun/16
Find a solution of the form ∅(x)=x^r Σ_(k=0) ^α c_k x^k (x > 0) for xy^(′′) + y′ − y = 0.
$${Find}\:{a}\:{solution}\:{of}\:{the}\:{form}\:\emptyset\left({x}\right)={x}^{{r}} \underset{{k}=\mathrm{0}} {\overset{\alpha} {\sum}}{c}_{{k}} {x}^{{k}} \:\left({x}\:>\:\mathrm{0}\right)\:{for}\: \\ $$$${xy}^{''} +\:{y}'\:−\:{y}\:=\:\mathrm{0}. \\ $$
Commented by Yozzii last updated on 29/Jun/16
∅(x)=Σ_(k=0) ^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) ∅′(x)=Σ_(k=0) ^α c_k (k+r)x^(k+r−1) ∅′(x)=c_0 rx^(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) ∅′(x)−∅(x)=c_0 rx^(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) ∅′(x)−∅(x)=c_0 rx^(r−1) −c_α x^(r+α) +Σ_(k=1) ^α (c_k (r+k)−c_(k−1) )x^(r+k−1) x∅′′(x)=Σ_(k=0) ^α c_k (k+r)(k+r−1)x^(k+r−1) x∅′′(x)=c_0 r(r−1)x^(r−1) +c_1 (r+1)rx^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∅′′(x)=c_0 r(r−1)x^(r−1) +Σ_(k=1) ^α c_k (k+r)(k+r−1)x^(k+r−1) x∅′′(x)+∅′(x)−∅(x)=0 ∴ x^(r−1) (c_0 r^2 −c_0 r+c_0 r)−c_α x^(r+α) +Σ_(k=1) ^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) +Σ_(k=1) ^α x^(k+r−1) (c_k (k+r)−c_(k−1) )=0 ⇒c_0 =c_α =0 ∴c_1 (1+r)−c_0 =0⇒c_1 =0 c_2 (2+r)−c_1 =0⇒c_2 =0 c_3 (3+r)−c_2 =0⇒c_3 =0 ∴ c_k =0 ∀k∈Z^≥ ∴ ∅(x)=0+0+0+0+...+0=0
$$\emptyset\left({x}\right)=\underset{{k}=\mathrm{0}} {\overset{\alpha} {\sum}}{c}_{{k}} {x}^{{k}+{r}} ={c}_{\mathrm{0}} {x}^{{r}} +{c}_{\mathrm{1}} {x}^{{r}+\mathrm{1}} +{c}_{\mathrm{2}} {x}^{{r}+\mathrm{2}} +…+{c}_{{k}} {x}^{{k}+{r}} +…+{c}_{\alpha} {x}^{\alpha+{r}} \\ $$$$\emptyset'\left({x}\right)=\underset{{k}=\mathrm{0}} {\overset{\alpha} {\sum}}{c}_{{k}} \left({k}+{r}\right){x}^{{k}+{r}−\mathrm{1}} \\ $$$$\emptyset'\left({x}\right)={c}_{\mathrm{0}} {rx}^{{r}−\mathrm{1}} +{c}_{\mathrm{1}} \left({r}+\mathrm{1}\right){x}^{{r}} +{c}_{\mathrm{2}} \left({r}+\mathrm{2}\right){x}^{{r}+\mathrm{1}} +{c}_{\mathrm{3}} \left({r}+\mathrm{3}\right){x}^{{r}+\mathrm{2}} +…+{c}_{{k}} \left({r}+{k}\right){x}^{{k}+{r}−\mathrm{1}} +…+{c}_{\alpha} \left({r}+\alpha\right){x}^{\alpha+{r}−\mathrm{1}} \\ $$$$\emptyset'\left({x}\right)−\emptyset\left({x}\right)={c}_{\mathrm{0}} {rx}^{{r}−\mathrm{1}} +\left({c}_{\mathrm{1}} \left({r}+\mathrm{1}\right)−{c}_{\mathrm{0}} \right){x}^{{r}} +\left({c}_{\mathrm{2}} \left({r}+\mathrm{2}\right)−{c}_{\mathrm{1}} \right){x}^{{r}+\mathrm{1}} +…+\left({c}_{{k}} \left({r}+{k}\right)−{c}_{{k}−\mathrm{1}} \right){x}^{{r}+{k}−\mathrm{1}} +…+\left({c}_{\alpha} \left({r}+\alpha\right)−{c}_{\alpha−\mathrm{1}} \right){x}^{{r}+\alpha−\mathrm{1}} −{c}_{\alpha} {x}^{{r}+\alpha} \\ $$$$\emptyset'\left({x}\right)−\emptyset\left({x}\right)={c}_{\mathrm{0}} {rx}^{{r}−\mathrm{1}} −{c}_{\alpha} {x}^{{r}+\alpha} +\underset{{k}=\mathrm{1}} {\overset{\alpha} {\sum}}\left({c}_{{k}} \left({r}+{k}\right)−{c}_{{k}−\mathrm{1}} \right){x}^{{r}+{k}−\mathrm{1}} \\ $$$${x}\emptyset''\left({x}\right)=\underset{{k}=\mathrm{0}} {\overset{\alpha} {\sum}}{c}_{{k}} \left({k}+{r}\right)\left({k}+{r}−\mathrm{1}\right){x}^{{k}+{r}−\mathrm{1}} \\ $$$${x}\emptyset''\left({x}\right)={c}_{\mathrm{0}} {r}\left({r}−\mathrm{1}\right){x}^{{r}−\mathrm{1}} +{c}_{\mathrm{1}} \left({r}+\mathrm{1}\right){rx}^{{r}} +{c}_{\mathrm{2}} \left({r}+\mathrm{2}\right)\left({r}+\mathrm{1}\right){x}^{{r}+\mathrm{1}} +{c}_{\mathrm{3}} \left({r}+\mathrm{3}\right)\left({r}+\mathrm{2}\right){x}^{{r}+\mathrm{2}} +…+{c}_{{k}} \left({k}+{r}\right)\left({k}+{r}−\mathrm{1}\right){x}^{{k}+{r}−\mathrm{1}} +…+{c}_{\alpha} \left({r}+\alpha\right)\left(\alpha+{r}−\mathrm{1}\right){x}^{\alpha+{r}−\mathrm{1}} \\ $$$${x}\emptyset''\left({x}\right)={c}_{\mathrm{0}} {r}\left({r}−\mathrm{1}\right){x}^{{r}−\mathrm{1}} +\underset{{k}=\mathrm{1}} {\overset{\alpha} {\sum}}{c}_{{k}} \left({k}+{r}\right)\left({k}+{r}−\mathrm{1}\right){x}^{{k}+{r}−\mathrm{1}} \\ $$$${x}\emptyset''\left({x}\right)+\emptyset'\left({x}\right)−\emptyset\left({x}\right)=\mathrm{0} \\ $$$$\therefore\:{x}^{{r}−\mathrm{1}} \left({c}_{\mathrm{0}} {r}^{\mathrm{2}} −{c}_{\mathrm{0}} {r}+{c}_{\mathrm{0}} {r}\right)−{c}_{\alpha} {x}^{{r}+\alpha} +\underset{{k}=\mathrm{1}} {\overset{\alpha} {\sum}}{x}^{{k}+{r}−\mathrm{1}} \left({c}_{{k}} \left({k}+{r}\right)^{\mathrm{2}} −{c}_{{k}−\mathrm{1}} \right)=\mathrm{0} \\ $$$${c}_{\mathrm{0}} {r}^{\mathrm{2}} {x}^{{r}−\mathrm{1}} −{c}_{\alpha} {x}^{{r}+\alpha} +\underset{{k}=\mathrm{1}} {\overset{\alpha} {\sum}}{x}^{{k}+{r}−\mathrm{1}} \left({c}_{{k}} \left({k}+{r}\right)−{c}_{{k}−\mathrm{1}} \right)=\mathrm{0} \\ $$$$\Rightarrow{c}_{\mathrm{0}} ={c}_{\alpha} =\mathrm{0} \\ $$$$\therefore{c}_{\mathrm{1}} \left(\mathrm{1}+{r}\right)−{c}_{\mathrm{0}} =\mathrm{0}\Rightarrow{c}_{\mathrm{1}} =\mathrm{0} \\ $$$${c}_{\mathrm{2}} \left(\mathrm{2}+{r}\right)−{c}_{\mathrm{1}} =\mathrm{0}\Rightarrow{c}_{\mathrm{2}} =\mathrm{0} \\ $$$${c}_{\mathrm{3}} \left(\mathrm{3}+{r}\right)−{c}_{\mathrm{2}} =\mathrm{0}\Rightarrow{c}_{\mathrm{3}} =\mathrm{0} \\ $$$$\therefore\:{c}_{{k}} =\mathrm{0}\:\forall{k}\in\mathbb{Z}^{\geqslant} \\ $$$$\therefore\:\emptyset\left({x}\right)=\mathrm{0}+\mathrm{0}+\mathrm{0}+\mathrm{0}+…+\mathrm{0}=\mathrm{0} \\ $$
Commented by nburiburu last updated on 29/Jun/16
xy′′+y′=(xy′)′=y if y=x^w ⇒ (w)^2 .x^(w−1) =x^w which is impossible for w∈R so the only way to continue would be to assume w=w(x) y=x^(w(x)) ⇒(w.x^w )′=x^w [w′(1+ln x)+w^2 /x]=x^w w′(1+lnx)+(w^2 /x)=1 w′(x+xlnx)+w^2 =x homogeneus solution: w′(x+xlnx)+w^2 =0 (1/w_h ^2 ) dw_h = ((−1)/(x(1+lnx)))dx −(1/w_h ) = − ln(1+lnx) w_h =(1/(ln(1+lnx))) Note: for the eq to have the structure mention as φ(x) the eq should be x^2 y′′+xy′−y=0
$${xy}''+{y}'=\left({xy}'\right)'={y} \\ $$$${if}\:{y}={x}^{{w}} \:\Rightarrow\:\left({w}\right)^{\mathrm{2}} .{x}^{{w}−\mathrm{1}} ={x}^{{w}} \:{which}\:{is}\:{impossible}\:{for}\:{w}\in\mathbb{R} \\ $$$${so}\:{the}\:{only}\:{way}\:{to}\:{continue}\:{would}\:{be} \\ $$$${to}\:{assume}\:{w}={w}\left({x}\right) \\ $$$${y}={x}^{{w}\left({x}\right)} \Rightarrow\left({w}.{x}^{{w}} \right)'={x}^{{w}} \left[{w}'\left(\mathrm{1}+{ln}\:{x}\right)+{w}^{\mathrm{2}} /{x}\right]={x}^{{w}} \\ $$$${w}'\left(\mathrm{1}+{lnx}\right)+\frac{{w}^{\mathrm{2}} }{{x}}=\mathrm{1} \\ $$$${w}'\left({x}+{xlnx}\right)+{w}^{\mathrm{2}} ={x} \\ $$$${homogeneus}\:{solution}: \\ $$$${w}'\left({x}+{xlnx}\right)+{w}^{\mathrm{2}} =\mathrm{0} \\ $$$$\frac{\mathrm{1}}{{w}_{{h}} ^{\mathrm{2}} }\:{dw}_{{h}} \:=\:\frac{−\mathrm{1}}{{x}\left(\mathrm{1}+{lnx}\right)}{dx} \\ $$$$−\frac{\mathrm{1}}{{w}_{{h}} }\:=\:−\:{ln}\left(\mathrm{1}+{lnx}\right) \\ $$$${w}_{{h}} =\frac{\mathrm{1}}{{ln}\left(\mathrm{1}+{lnx}\right)} \\ $$$$ \\ $$$${Note}:\:{for}\:{the}\:{eq}\:{to}\:{have}\:{the}\:{structure} \\ $$$${mention}\:{as}\:\phi\left({x}\right)\:{the}\:{eq}\:{should}\:{be} \\ $$$${x}^{\mathrm{2}} {y}''+{xy}'−{y}=\mathrm{0} \\ $$$$ \\ $$$$ \\ $$
Commented by WAI LIN last updated on 01/Jul/16
Using the power series method(Frobenius method)
$${Using}\:{the}\:{power}\:{series}\:{method}\left({Frobenius}\:{method}\right) \\ $$

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