Question

Find the general solution of xy"+(2-2x)y'+(x-2)y=0, given that y1=e^x satisfies the complementary equation.

Answer 1

y=ue^x
y'=u'e^(x)+ue^(x)
y"=u"e^(x)+2u'e^(x)+ue^(x)
Subbing:
xy"+(2-2x)y'+(x-2)y=u"xe^(x)+2u'e^(x)=0
\[u''+\frac{ 2 }{ x }u'=0\]
z=u' Subbing:
\[z'+\frac{ 2 }{ x }z=0\]
the integrating factor is x^2
\[x^2(z'+\frac{ 2 }{ x }z)=0\]
\[\frac{ d }{ dx }(x^2z)=0\]
\[x^2z=C\]
\[z=u'=\frac{ C }{ x^2 }\]

Answer 2

integrate,
\[u=-\frac{ C _{1} }{ x }+C _{2}\]
So
\[y=e ^{x}(-\frac{ C _{1} }{ x }+C _{2})\]

Answer 3

Is my answer right? Can anyone confirm?

Answer 4

@Kainui @Luigi0210 @Preetha

Answer 5

What do you get when you plug it into the differential equation?