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

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More answers

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

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