Question

Can someone help me? The answer that I get does not equate to 0...
This is the question.
Given that y=2x+1/(x) show that (x)*d^2y/dx^2 +(2)*dy/dx=0

Answer 1

Concerns:
1. is it y = 2x + 1/x , as you typed it, or is it y = (2x+1)/x

2. are we showing that
x(the second derivative) + 2(first derivative) = 0 ?

Answer 2

doyoumean
y = (2x+1)/(x) ?
or
y = 2 x + (1/x) ?
if the second
dy /dx = 2 - 1/x^2
d^2 y /dx^2 = 2x/x^4 = 2/x^3
then
(x)*d^2y/dx^2 +(2)*dy/dx
= 2/x^2 +2(2-1/x^2)
= 4

Answer 3

Assuming my second guess:

y' = (2x - 2x - 1)/x^2
= -1/x^2 or - x^-2
y'' = 2x^-3

LS = x(2/x^3) + 2(-x^-2)
= 2/x^2 - 2/x^2
= 0
= RS

Answer 4

if you mean
y = (2x+1)/(x)
dy/dx = (2x - (2x+1) ) / x^2 = -1/x^2
d^2y/dx^2 = 1(2x) / x^4 = 2/x^3
then
(x)*d^2y/dx^2 +(2)*dy/dx
= 2/x^2 - 2/x^2 = 0

Answer 5

it was (2x+1)/x

Answer 6

We all know that now :)

Answer 7

@mathhelper
1. it was (2x+1)/x
2. yes that's what it was supposed to show

Answer 8

the solution of xy" + 2y' = 0 is
y = c1 + c2/x
so y=(2x+1)/x = 2 + 1/x
will certainly work