Question

Solve the following eqn

Answer 1

\(x\frac{dy}{dx}=cot~y\)

Answer 2

do you know how to do this one?

Answer 3

just rewrite as
\(1/x \text{ dx}=\tan y \text{ dy}\)
integrate
\(\int dx/x=\int y dy\)
super easy

Answer 4

Yes,
I got
\(y=cos(x+c)\)
is it correct?

Answer 5

one sec let me check

Answer 6

I got something else, can you show what you did?

Answer 7

\(-In|cos~y|=In~x+c\)
\(In|cos~y|^{-1}=In~x+c\)----> not really sure if we can do like this XD
\(cos^{-1}y=x+c\)
\(y=cos(x+c)\)

Answer 8

should be -ln cos y = ln x + C
ln (sec y) = ln x + C
sec y = C*e^x

Answer 9

owh shouldn't it be
\(In(sec~y)=Inx +C\)
\(sec~y=x+e^C\)
\(y=sec^{-1}(x+e^C)\)

Answer 10

ln(sec y)=ln x +C
e^(ln sec y)=e^(ln x +C)
sec y = e^(ln x)*e^C
sec y =C*e^ln(x) -> call e^C just C
sec y = C*x
You don't want to use the inverse trig function because that's not a real inverse, it will remove solutions from the relation that satisfies the equation I think

Answer 11

owh I c
if we integrate \(\frac{1}{x}dx\)
would it become
\(In(x+C)\) or \(In(x)+C\)

Answer 12

ln(x) + C
the constant is always outside...

Answer 13

just like if you integrate cos(x) you get sin(x)+C and not sin(x+C)

Answer 14

http://prntscr.com/m3awu8
shouldnt it be
+ ?

Answer 15

because if you derive ln(x+C) you get 1/(x+c)

Answer 16

no, read through it,
e^(ln x +C)=e^(ln x)*e^(C)

Answer 17

exponent law

Answer 18

\(\color{#0cbb34}{\text{Originally Posted by}}\) @nuts
because if you derive ln(x+C) you get 1/(x+c)
\(\color{#0cbb34}{\text{End of Quote}}\)
owh,yes,dats right
Thanks for clearing my thoughts XD

Answer 19

\(\color{#0cbb34}{\text{Originally Posted by}}\) @nuts
no, read through it,
e^(ln x +C)=e^(ln x)*e^(C)
\(\color{#0cbb34}{\text{End of Quote}}\)
owhhh, I c
LOL
*facepalm*
Aite thanks bro

Answer 20

yup no worries man, let me know if you have other questions!

Answer 21

Alright,will do! :D