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(adsbygoogle = window.adsbygoogle || []).push({});

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) + c3ln xcos(lnx) + c4ln xsin (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

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# Euler Equation please help

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