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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