Question

Find the general solution of the differential equation if x is restricted to the interval (0, infinity). (x^2)y''-2xy'-4y=0

Answer 1

That's called cauchy-euler

(r^2-r)-2(r)-4=0

solve for r

r=-1, 4

Answer 2

y= C t^-1 + C t^4

Answer 3

t should be x

Answer 4

Ok I think I get it thanks! The only thing, why is it (r^2-r)?

Answer 5

we initially assume that the solution must be

y= t^r

y'= r t^(r-1)

y''= r(r-1) t^(r-2)

Answer 6

Ooooo ok that makes sense. Thanks a lot!

Answer 7

Substitute \(x=e^t \implies t=\ln{x}\), then:
\[\frac{dy}{dx}={1 \over x}{dy \over dt} \text{ and } {d^2y \over dx^2}={1 \over x^2}({d^2y \over dt^2}-{dy \over dy})\]
Plug this into your original DE you get:
\[y''(t)-y'(t)-2y''(t)-4y(t)=0 \implies y''(t)-3y'(t)-4y(t)=0.\]
The auxiliary equation of this HDE is \(m^2- 3m-4=(m-4)(m+1)=0\).
Then the solution is \(y(t)=c_1e^{4t}+c_2e^{-t}\).
Re-substitute with \(t=\ln{x}\) and that should give you the general solution:
\[y(x)=c_1x^4+c_2x^{-1}.\]

Answer 8

imrans solution seems easier.

Answer 9

I get this one as well. Thanks a lot for your help. Now I'm gonna refer to this and try to do similar problems. Thanks!