Question

Find all solutions of y''+(2/x)y'=1 ; x>0

Answer 1

ok. so first step simplify.

Answer 2

see ya.

Answer 3

@phi @amistre64 @experimentX

Answer 4

Any hints on how to solve it ?

Answer 5

No, just the problem

Answer 6

I got it! On a side note, do you know how to solve complex roots of characteristic equations such as:
(t^2)y''-(4t)y'+6y=0 using Euler equations?

Answer 7

What is the solution?

Answer 8

or, the approach ?

Answer 9

I looked at how Wolfram alpha does it. They substitute v=y' and v'=y'' then is just becomes a typical equation, but the answer it gives has two constants and I wasn't sure if that was correct

Answer 10

yes, it (finally) came to me. find the integrating factor in the usual way, it is x^2

x^2 y'' + 2x y' = x^2

we know d/dx (x^2 y') = x^2 d/dx y' + y' d/dx(x^2) = x^2 y'' + y' 2x
the left side is an exact differential:

d (x^2 y') = x^2 dx
integrate both sides
x^2 y' = x^3/3 + C
y' = x/3 + C x^-2

integrate again:
y = x^2/6 + - C/x + D
C is an unknown constant, so we could just as well write

y= x^2/6 + C/x + D