y''+2y'+y=(e^-x)/(x^4) Solve using variation of parameters.
r^2 +2r +1
r=-1,-1 So, yc(x)= (C1)e^-x+(C2)e^-x(x)
y1=e^-x and y2=(e^-x)(x)
W = le^-x e^-x(x)l
l-e^-x e^-x + -e^-x(x)l
=e^-x^2 (2-x)
then, e^-x Integral (e^-x)e^-x(x)/x^4(e^-x^2(2-x)
Im not sure if this is correct though : \ Is there an easier way for this method?


LinkBack URL
About LinkBacks