# Thread: cauchy uler differential equation: y" + P(x) y' + Z(x) y = 0

1. ## cauchy uler differential equation: y" + P(x) y' + Z(x) y = 0

y"+P(x)y'+Z(x)y=0
is second order homogeneous differential equation. if yp is particular solution then y=yp.u, we can transform u into first order differential equation.
If m(m-1)+mxP(x)+Q(x)x2 =0, yparticular=xm
if m2+mP(x)+Q(x)=0, yparticular=emx
question: 1.(x-1)y"-xy'+y=0, find y

the answer on the book said that m2(x-1)-mx+1=(m-1)(mx-m-1)=0. m=1 and yp= ex can someone explain how to get this?? thanks!

2. Originally Posted by devinamuljono
y"+P(x)y'+Z(x)y=0
is second order homogeneous differential equation. if yp is particular solution then y=yp.u, we can transform u into first order differential equation.
If m(m-1)+mxP(x)+Q(x)x2 =0, yparticular=xm
if m2+mP(x)+Q(x)=0, yparticular=emx
question: 1.(x-1)y"-xy'+y=0, find y

the answer on the book said that m2(x-1)-mx+1=(m-1)(mx-m-1)=0. m=1 and yp= ex can someone explain how to get this?? thanks!
What are your thoughts?

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3. Originally Posted by devinamuljono
y"+P(x)y'+Z(x)y=0
is second order homogeneous differential equation. if yp is particular solution then y=yp.u, we can transform u into first order differential equation.
If m(m-1)+mxP(x)+Q(x)x2 =0, yparticular=xm
if m2+mP(x)+Q(x)=0, yparticular=emx
question: 1.(x-1)y"-xy'+y=0, find y

the answer on the book said that m2(x-1)-mx+1=(m-1)(mx-m-1)=0. m=1 and yp= ex can someone explain how to get this?? thanks!
Please specify at what point the book's explanation stopped making sense. Thank you!

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