- #1
Beretta
- 39
- 0
I did solve this differential equation (x^4)y'''' + 6x^3y''' + 9x^2 + 3xy' + y = 0 using Cauchy Euler Equation. I got X^m (m^2 + 1)^2 = 0
I'm not sure how to get the roots of (m^2 + 1)^2. In my calculation I got
m = -i, +i, -i, +i when I put m^2 = -1. In the book they have m = (+-) (squar(-4))/2 = (+-)2i. I don't know how they get this result. Any way, according to their result:
y = c1 cos(in x) + c2 sin (lnx) + c3 ln x cos(lnx) + c4 ln x sin (lnx).
According to mine:
y = c1 cos(in x) + c2 sin (lnx) + c3 cos(lnx) + c4 sin (lnx).
Where did they get the ln x before the last cos(ln x) and sin(ln x) from?
Thank you for your help
I'm not sure how to get the roots of (m^2 + 1)^2. In my calculation I got
m = -i, +i, -i, +i when I put m^2 = -1. In the book they have m = (+-) (squar(-4))/2 = (+-)2i. I don't know how they get this result. Any way, according to their result:
y = c1 cos(in x) + c2 sin (lnx) + c3 ln x cos(lnx) + c4 ln x sin (lnx).
According to mine:
y = c1 cos(in x) + c2 sin (lnx) + c3 cos(lnx) + c4 sin (lnx).
Where did they get the ln x before the last cos(ln x) and sin(ln x) from?
Thank you for your help