Taller ecuaciones

Cristian Quintero Castañeda 2009217058
Gustavo Vergel 2005211015

2. y''+a2y=f(x) D2+a2y=fx D2+a2y=0
La ecuación auxiliar es m2+a2=0 m=±aiyh=c1cosax+c2sinax
wy1,y2=cos⁡(ax)sin⁡(ax)-asin⁡(ax)acos⁡(ax)=cos2ax+sin2ax=a
u1'=0sin⁡(ax)f(x)cos⁡(ax)w(y1,y2)=-fxsin⁡(ax)a=-fxsin⁡(ax)
u1=-fxsinaxdx=-1afxsinaxdx =-1afμsinaμdμu2'=cos⁡(ax)0-sin⁡(ax)f(X)w(y1,y2)=-fxcos⁡(ax)1=fxcos⁡(ax)
u2=fxcosaxdx=1afxcosaxdx =1afμcosaμdμ
yp=-1acosax0xfμsinaμdμ+1asen(ax)0xfμcosaμdμyp=1a0xfμ(sinaμ*cosax-cosax*sin⁡(aμ))dμ
yp=1a0xfμ(sin(ax-μ))dμ
Luego, la solución general es:
y=c1cosax+c2sinax+1a0xfμ(sin(ax-μ))dμ
Como y0=y'0=0
0=ac1 ; c1=0
y=c2sinax+1a0xfμ(sin(ax-μ))dμy'=c2cosax-1a20xfμ(sin(ax-μ))dμ+1afμ(sin(ax-μ))dμddμ
y=1a0xfμ(sin(ax-μ))dμ

3.
x3y'''+xy'-y=xlnx
sea t=lnx ; x= et
dydx=dydt*dtdx ; dydx=dydt*1x ; dydt=xdydt; d2ydx2=1x*ddxdydt-1x2*dydt
d2ydx2=1x*d2ydt21x-1x2dydt ; x2d2ydx2=d2ydx2-dydt
d3ydx3=ddx1x2d2ydt2-dydt+1x2d3ydt3dtdx-d2ydt2dtdxd3ydx3=-2x3d2ydt2-dydt+1x2d3ydt31x-d2ydt21x
x3d3ydx3=d3ydt3-3d2ydt2+2dydt
reemplazando en ; x3y'''+xy'-y=xlnx
d3ydt3-3d2ydt2+2dydt+dydt-y=ett
D3-3D2+3D-1y=tet ; D-13y=tet ; D-13y=0 ; m1=m2=m3=1yc=c1+c2t+c3t2et
ft=tet es anulado por D-12 ; D-12ft=0 ; m-12=0 ; m4=m5=1
yp=t3A+Btet
yp'=3t2A+Btet+Bt3et+t3(4+Bt)etyp''=6tA+Btet+6Bt2et+6t2A+Btet+2Bt3et+t3(A+Bt)et
yp'''=6A+Btet+6Bt2et+6tA+Btet+12Btet+6Bt2et+12tA+Btet+6Bt2et+6t2A+Btet+6Bt2et+2Bt3et+3t2A+Btet+Bt3et+t3(A+Bt)et
yp'''-3yp''+3yp'-yp=tet
6A+6Bt+18Btet=tet
6A=0; A=0 ; 24B=1 ; B=124
yp=t30+124tet ; yp=124t4et
y=c1+c2t+c3t2+124t4et
y=xc1+c2(lnx)+c3(lnx)2+124x(lnx)4
y=c1x+c2xlnx+c3x(lnx)2+124x(lnx)4