- #1
iScience
- 466
- 5
problem: xy''-4y'=x^4
when i solve it i get y(homogeneous)=C1+C2x^5 which is fine because that's what the back of the book says
but for the particular solution, i get .. C1=-(1/25)x^5 , C2=(1/5)lnx
so y(general)= C1+C2x^5-(1/25)x^5+(1/5)lnx(x^5)
but wolframalpha as well as the back of my book both say the answer is...
y(general)=C1+C2x^5+(1/5)lnx(x^5)
where does the -(1/25)x^5 term go?...
when i solve it i get y(homogeneous)=C1+C2x^5 which is fine because that's what the back of the book says
but for the particular solution, i get .. C1=-(1/25)x^5 , C2=(1/5)lnx
so y(general)= C1+C2x^5-(1/25)x^5+(1/5)lnx(x^5)
but wolframalpha as well as the back of my book both say the answer is...
y(general)=C1+C2x^5+(1/5)lnx(x^5)
where does the -(1/25)x^5 term go?...