Question

\[{dy \over dx}= 4{y^2 \over x^2}+5{y \over x} +1\]

Answer 1

let \[y=vx\]\[{dy \over dx}=x{dv \over dx}+v\]

Answer 2

yeah \[y=y(x)\]

Answer 3

\[x{dv \over dx}+v= 4v^2+5v+1\]

Answer 4

\[{dv \over dx}={4v^2+4v+1 \over x}\]
\[\int{dv \over 4v^2+4v+1}= \int{dx\over x}\]

Answer 5

am i doing this right?

Answer 6

yes

Answer 7

how am i supposed to integrate the LHS

Answer 8

complete the square, it will be an arctan thing..

Answer 9

maybe i shouldn't have misused the v

Answer 10

completing the square

4v^2+4v+1

a^2 +2ab+b^2

2(2)(1)=4 so b must be 1
b^2=1

this is perfect square
(2v+1)(2v +1)=4v^2+4v+1

Answer 11

...so partial fractions works too then I guess

Answer 12

\[\int{dv \over (2v+1)^2}=\int {dx \over x}\]\[{-1 \over 4v+2}=ln(x) +c_1\]\[4v+2 = ln(1/x)+c_2\]

Answer 13

haha, oh yeah....
I was overcomplicating it

Answer 14

\[y=x/4(c_3-lnx)\]

Answer 15

oh wait, the answer in my book has arctans and lns

Answer 16

i must have put a negative in the wrong stop somewhere

Answer 17

the answer in the back of my book is
\[tan^{-1}({y+3 \over x+2})+{1 \over 2}ln(x^2+y^24x+6y+13)=c\]

Answer 18

someone put me on the right track,

Answer 19

where did i go wrong

Answer 20

.. somebody?!