1. Dec 6, 2004

### Beretta

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?

2. Dec 6, 2004

### HallsofIvy

Staff Emeritus
That's very peculiar.

Yes, the "characteristic equation" reduces to (m2+ 1)2= 0 which has m= i and -i as double roots.

You say your text gives the roots as 2i and -2i but then the "2"s do not show up in the book's solution to the equation???

Remember how, in an equation with constant coefficients, if your characteristic equation had multiple roots, you had to multiply by "x" to get a new, independent solution? Same thing here except that it is ln x you multiply by.

The reason for that is this: if you let t= ln x, then dy/dx= (dy/dt)(dt/dx)= (1/x)dy/dt so that x dy/dx becomes just dy/dt. The same thing happens to the other derivatives so the the "Euler-type" equation becomes an equation with constant coefficients. That new equation turns out to have exactly the same characteristic equation, (m2+ 1)2= 0 so that the characteristic roots are still i, -i, i, -i. That means the general solution, in terms of t, is
y(t)= C1cos(t)+ C2sin(t)+ C3t cos(t)+ C4t sin(t).

Putting t= ln(x) back in,
y(x)= C1cos(ln(x))+ C2sin(lin(x))+ C3ln(x) cos(ln(x))+ C4ln(x) sin(ln(x)).

3. Dec 22, 2004

### noppakhuns

The four solutions have to be linearly independent and you may prove the solutions for characteristic equations with repeated roots by reducing the order. Set up the condition as you know the first solution for a homogeneous equation, then solve for another by using y2=u(x)y1, substute in and solve for u(x), you get a nice y2=xe^rx, where r is arepested root from characteristic equation.