(1−x^2 )(d^2 y/dx^2 ) −2x(dy/dx) + p(p+1)y = 0 in descending power of x. what is the solution?


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Question Number 83327 by john santu last updated on 01/Mar/20

$(1 - x^{2}) \frac{d^{2} y}{{dx}^{2}} - 2 x \frac{dy}{dx} + p (p + 1) y = 0$ $in descending power of x . what is$ $the solution ?$
Commented by Joel578 last updated on 01/Mar/20

$It is {Legendre}^{'} s equation .$ $Its solutions known as Legendre polynomial$
Commented by niroj last updated on 01/Mar/20
$(1−x^2 )(d^2 y/dx^2 )−2x(dy/dx)+p(p+1)y=0....(i) Sol^n : let, x=0 , f(x)≠0 i.e. x=0 is ordinary point i.e. solution of equation is y= Σ_(r=0) ^∞ a_r x^r (dy/dx)= Σ_(r=0) ^∞ a_r rx^(r−1) (d^2 y/dx^2 ) = Σ_(r=0) ^∞ a_r r(r−1)x^(r−2) Put, the value of y,y_1 and y_2 in eq^n (i) (1−x^2 )Σ_(r=0) ^∞ a_r r(r−1)x^(r−2) −2xΣ_(r=0) ^∞ a_r rx^(r−1) +p(p+1)Σ_(r=0) ^∞ a_r x^r =0 or, Σ_(r=0) ^∞ a_r r(r−1)x^(r−2) −Σ_(r=0) ^∞ a_r r(r−1)x^r −2Σ_(r=0) ^∞ a_r rx^r +p(p+1)Σ_(r=0) ^∞ a_r x^r =0 or, Σ_(r=0) ^∞ a_r r(r−1)x^(r−2) −Σ_(r=0) ^∞ a_r [r(r−1)+2r−p(p+1)]x^r =0 or_, Σ_(r=0) ^∞ a_r r(r−1)x^(r−2) −Σ_(r=0) ^∞ a_r [r^2 +r−p(p+1)]x^r =0 or_(, ) Σ_(r=0) ^∞ a_r [(r^2 −r)x^(r−2) −(r^2 +r−p^2 −p)]x^r =0 or,a_r r(r−1)x^(r−2) −a_r [r(r+1)−p(p+1)]x^r =0 put r=r+2 a_(r+2) (r+2)(r+1)x^r −a_r [(r+2)(r+3)−p(p+1)]x^r =0 or, [a_(r+2) (r+2)(r+1)−a_r {(r+2)(r+3)−p(p+1)}]x^r =0 or, a_(r+2) (r+2)(r+1)−a_r {(r+2)(r+3)−p(p+1)}=0 a_(r+2) = (((r+2)(r+3)−p(p+1))/((r+2)(r+1)))a_r put r=0, a_2 = ((2.3−p(p+1))/(2.1))a_0 then put descending value of r in terms of power series y= Σ_(r=0) ^∞ a_r x^r = a_0 +ax+a_2 x^2 +....$
$(1 - x^{2}) \frac{d^{2} y}{{dx}^{2}} - 2 x \frac{dy}{dx} + p (p + 1) y = 0 . . . . (i)$ ${Sol}^{n} :$ $let, x = 0, f (x) \neq 0$ $i . e . x = 0 is ordinary point$ $i . e . solution of equation is$ $y = \underset{r = 0}{\sum^{\infty}} a_{r} x^{r}$ $\frac{dy}{dx} = \underset{r = 0}{\sum^{\infty}} a_{r} {rx}^{r - 1}$ $\frac{d^{2} y}{{dx}^{2}} = \underset{r = 0}{\sum^{\infty}} a_{r} r (r - 1) x^{r - 2}$ $Put, the value of y, y_{1} and y_{2} in {eq}^{n} (i)$ $(1 - x^{2}) \underset{r = 0}{\sum^{\infty}} a_{r} r (r - 1) x^{r - 2} - 2 x \underset{r = 0}{\sum^{\infty}} a_{r} {rx}^{r - 1} + p (p + 1) \underset{r = 0}{\sum^{\infty}} a_{r} x^{r} = 0$ $or, \underset{r = 0}{\sum^{\infty}} a_{r} r (r - 1) x^{r - 2} - \underset{r = 0}{\sum^{\infty}} a_{r} r (r - 1) x^{r} - 2 \underset{r = 0}{\sum^{\infty}} a_{r} {rx}^{r} + p (p + 1) \underset{r = 0}{\sum^{\infty}} a_{r} x^{r} = 0$ $or, \underset{r = 0}{\sum^{\infty}} a_{r} r (r - 1) x^{r - 2} - \underset{r = 0}{\sum^{\infty}} a_{r} [r (r - 1) + 2 r - p (p + 1)] x^{r} = 0$ ${or}_{,} \underset{r = 0}{\sum^{\infty}} a_{r} r (r - 1) x^{r - 2} - \underset{r = 0}{\sum^{\infty}} a_{r} [r^{2} + r - p (p + 1)] x^{r} = 0$ ${or}_{,} \underset{r = 0}{\sum^{\infty}} a_{r} [(r^{2} - r) x^{r - 2} - (r^{2} + r - p^{2} - p)] x^{r} = 0$ $or, a_{r} r (r - 1) x^{r - 2} - a_{r} [r (r + 1) - p (p + 1)] x^{r} = 0$ $put r = r + 2$ $a_{r + 2} (r + 2) (r + 1) x^{r} - a_{r} [(r + 2) (r + 3) - p (p + 1)] x^{r} = 0$ $or, [a_{r + 2} (r + 2) (r + 1) - a_{r} {(r + 2) (r + 3) - p (p + 1)}] x^{r} = 0$ $or, a_{r + 2} (r + 2) (r + 1) - a_{r} {(r + 2) (r + 3) - p (p + 1)} = 0$ $a_{r + 2} = \frac{(r + 2) (r + 3) - p (p + 1)}{(r + 2) (r + 1)} a_{r}$ $put r = 0, a_{2} = \frac{2.3 - p (p + 1)}{2.1} a_{0}$ $then put descending value of r in$ $terms of power series$ $y = \underset{r = 0}{\sum^{\infty}} a_{r} x^{r} = a_{0} + ax + a_{2} x^{2} + . . . .$
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