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

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

Mathematics
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let \[y=vx\]\[{dy \over dx}=x{dv \over dx}+v\]
yeah \[y=y(x)\]
\[x{dv \over dx}+v= 4v^2+5v+1\]

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Other answers:

\[{dv \over dx}={4v^2+4v+1 \over x}\] \[\int{dv \over 4v^2+4v+1}= \int{dx\over x}\]
am i doing this right?
yes
how am i supposed to integrate the LHS
complete the square, it will be an arctan thing..
maybe i shouldn't have misused the v
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
...so partial fractions works too then I guess
\[\int{dv \over (2v+1)^2}=\int {dx \over x}\]\[{-1 \over 4v+2}=ln(x) +c_1\]\[4v+2 = ln(1/x)+c_2\]
haha, oh yeah.... I was overcomplicating it
\[y=x/4(c_3-lnx)\]
oh wait, the answer in my book has arctans and lns
i must have put a negative in the wrong stop somewhere
the answer in the back of my book is \[tan^{-1}({y+3 \over x+2})+{1 \over 2}ln(x^2+y^24x+6y+13)=c\]
someone put me on the right track,
where did i go wrong
.. somebody?!

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