x2y′′+xy′+4y=2xlnx Homogeneous equation
x2y′′+xy′+4y=0
Let y=xr
y′=rxr−1
y′′=r(r−1)xr−2
r(r−1)+r+4=0
r=±2iThe general solution of the homogeneous equation is
yh=C1cos(2lnx)+C2sin(2lnx)
x2y′′+xy′+4y=2xlnx Divide both sides by x2
y′′+x1y′+x24y=x2lnx=>g(x)=x2lnx Wronskian
W(y1,y2)=∣∣cos(2lnx)−x2sin(2lnx)sin(2lnx)x2cos(2lnx)∣∣=x2
W1=∣∣0x2lnxsin(2lnx)x2cos(2lnx)∣∣=−x2lnxsin(2lnx)
W2=∣∣cos(2lnx)−x2sin(2lnx)0x2lnx∣∣=x2lnxcos(2lnx)
u1=∫WW1dx=∫x2−x2lnxsin(2lnx)dx
=−∫lnxsin(2lnx)dx
=−251x(5lnx+3)sin(2lnx)+251(10lnx−4)cos(2lnx)
u2=∫WW2dx=∫x2x2lnxcos(2lnx)dx
=∫lnxcos(2lnx)dx
=251x(10lnx−4)sin(2lnx)+251(5lnx+3)cos(2lnx)
Particular solution of the non-homogeneous equation is
yp=u1y1+u2y2
yp=−251x(5lnx+3)sin(2lnx)cos(2lnx)
+251(10lnx−4)cos2(2lnx)
+251x(10lnx−4)sin2(2lnx)
+251(5lnx+3)sin(2lnx)cos(2lnx) The general solution of the homogeneous equation is
y=C1cos(2lnx)+C2sin(2lnx)
−251x(5lnx+3)sin(2lnx)cos(2lnx)
+251(10lnx−4)cos2(2lnx)
+251x(10lnx−4)sin2(2lnx)
+251(5lnx+3)sin(2lnx)cos(2lnx)
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