Question

Check if the given function is a solution to the ODE

Answer 1

\[xy \prime +y=0 , y(x)=\frac{ 1 }{ x }\]

Answer 2

\[y'=\frac{-1}{x^2}\]

Answer 3

So \[xy'+y = x\frac{-1}{x^2}+\frac{1}{x}\]

Answer 4

=1/x + (-1/x)

Answer 5

that indeed equals zero so it works

Answer 6

Confused one where the x came from in front of -1/x^2

Answer 7

what x :O

Answer 8

u have
y=1/x
and ODE of
xy'+y=0
first find y' and apply it in the ODE :)

Answer 9

\[xy \prime +y=>x< \frac{ -1 }{ x^2 }+\frac{ 1 }{ x }\]

Answer 10

from where `<` came from ?

Answer 11

y'=-1/x^2 right ?

Answer 12

yess so when you put it in the ODE why is there an x in front of it? I get the 1/x

Answer 13

the ODE is
xy'+y=0
so
x(-1/x^2)+1/x =0
which is correct :D thats it

Answer 14

OH so basically you just sub in y and y prime? I'm trying to brush up on this lol

Answer 15

The x came form multiplying by x in the xy'+y=0

Answer 16

thank you :)

Answer 17

its in ur question :)
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