Using Cauchy-Euler equations?

Question

How would one find the general solution to a intial value problem using cauchy-euler equation.

(x^2)y" + 3xy' + 5y = 0 y(1)=1 y'(1)= -1

Answer 1

Let x=e^t. Then dy/dt = dy/dx dx/dt = e^t dy/dx, and
d^2 y/dt^2 = d(e^t dy/dx)/dt = e^t dy/dx + e^(2t) d^2 y/dx^2, so

dy/dx = e^-t dy/dt and d^2 y/dx^2 = e^(-2t) (d^2 y/dt^2 - dy/dt).

Therefore, the equation becomes

e^(2t)e^(-2t)(d^2 y/dt^2 - dy/dt) + 3e^t e^(-t) dy/dt + 5y = 0,
or
y'' + 2y' + 5y = 0,
where ' now denotes differentiation with respect to t, not x.

The characteristic equation is r^2 + 2r + 5 = 0, which has complex roots r = -1+2i, -1-2i. In general, the characeristic equation of x^2 + bxy + cy = 0 is r^2 + (b-1)r + c = 0.

Therefore, the general solution is

y = |x|^(-1) ( c1 cos(2 ln|x|) + c2 sin(2 ln|x|) )

where c1 and c2 are chosen so that the initial conditions are satisfied. In this case, you get c1 = 1 and c2 = -1.

In general, when the characteristic equation has two real roots u and v, the general solution is y = c1|x|^u + c2|x|^v.

If there is a double root u, the general solution is y = (c1 + c2 ln |x|)|x|^u.

If there are two complex roots u+iv, u-iv, the general solution is |x|^u(c1 cos(v ln|x|) + c2 sin(v ln|x|)).