Re: Help DSolve

*To*: mathgroup at smc.vnet.net*Subject*: [mg39173] Re: Help DSolve*From*: "J.L.Garrido" <garrido at ruth.upc.es>*Date*: Sun, 2 Feb 2003 01:13:22 -0500 (EST)*References*: <b1afku$pg9$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

En/Na gianpf ha escrit: > I want to solve : > > (1+x)y''[x]-2(x+2)y'[x]+4y[x) == x > > Could someone can find a way out for this ? > > Thanks. Jean, y1[x_] = 2x^2 + 6x + 5 y[x_]=(2x^2 + 6x + 5)*z[x] (1+x)y''[x]-2(x+2)y'[x]+4y[x]=(-8-14x-12x^2-4x^3) z'[x]+(5+11x+8x^2+2x^3)z''[x]=0 z[x]=1/4 E[2x]/(5+6x+2x^2) y2[x]=1/4 Exp[2x] y[x]=A y1[x]+B y2[x] (homogeneous solution) y[x]=1/2 x+1/2 (particular solution) y[x_]=A ( 2x^2 + 6x + 5 )+1/4 B Exp[2x]+1/2 x+1/2 (general solution) In[50]:= (1+x)y''[x]-2(x+2)y'[x]+4y[x]==x//FullSimplify Out[50]= True