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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.

Findasolutionoftheform(x)=xrαk=0ckxk(x>0)for xy+yy=0.

Commented byYozzii 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

(x)=αk=0ckxk+r=c0xr+c1xr+1+c2xr+2+...+ckxk+r+...+cαxα+r (x)=αk=0ck(k+r)xk+r1 (x)=c0rxr1+c1(r+1)xr+c2(r+2)xr+1+c3(r+3)xr+2+...+ck(r+k)xk+r1+...+cα(r+α)xα+r1 (x)(x)=c0rxr1+(c1(r+1)c0)xr+(c2(r+2)c1)xr+1+...+(ck(r+k)ck1)xr+k1+...+(cα(r+α)cα1)xr+α1cαxr+α (x)(x)=c0rxr1cαxr+α+αk=1(ck(r+k)ck1)xr+k1 x(x)=αk=0ck(k+r)(k+r1)xk+r1 x(x)=c0r(r1)xr1+c1(r+1)rxr+c2(r+2)(r+1)xr+1+c3(r+3)(r+2)xr+2+...+ck(k+r)(k+r1)xk+r1+...+cα(r+α)(α+r1)xα+r1 x(x)=c0r(r1)xr1+αk=1ck(k+r)(k+r1)xk+r1 x(x)+(x)(x)=0 xr1(c0r2c0r+c0r)cαxr+α+αk=1xk+r1(ck(k+r)2ck1)=0 c0r2xr1cαxr+α+αk=1xk+r1(ck(k+r)ck1)=0 c0=cα=0 c1(1+r)c0=0c1=0 c2(2+r)c1=0c2=0 c3(3+r)c2=0c3=0 ck=0kZ (x)=0+0+0+0+...+0=0

Commented bynburiburu 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=(xy)=y ify=xw(w)2.xw1=xwwhichisimpossibleforwR sotheonlywaytocontinuewouldbe toassumew=w(x) y=xw(x)(w.xw)=xw[w(1+lnx)+w2/x]=xw w(1+lnx)+w2x=1 w(x+xlnx)+w2=x homogeneussolution: w(x+xlnx)+w2=0 1wh2dwh=1x(1+lnx)dx 1wh=ln(1+lnx) wh=1ln(1+lnx) Note:fortheeqtohavethestructure mentionasϕ(x)theeqshouldbe x2y+xyy=0

Commented byWAI LIN last updated on 01/Jul/16

Using the power series method(Frobenius method)

Usingthepowerseriesmethod(Frobeniusmethod)

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