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

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UnkleRhaukus
 5 years ago
Best ResponseYou've already chosen the best response.0let \[y=vx\]\[{dy \over dx}=x{dv \over dx}+v\]

UnkleRhaukus
 5 years ago
Best ResponseYou've already chosen the best response.0\[x{dv \over dx}+v= 4v^2+5v+1\]

UnkleRhaukus
 5 years ago
Best ResponseYou've already chosen the best response.0\[{dv \over dx}={4v^2+4v+1 \over x}\] \[\int{dv \over 4v^2+4v+1}= \int{dx\over x}\]

UnkleRhaukus
 5 years ago
Best ResponseYou've already chosen the best response.0am i doing this right?

UnkleRhaukus
 5 years ago
Best ResponseYou've already chosen the best response.0how am i supposed to integrate the LHS

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.1complete the square, it will be an arctan thing..

UnkleRhaukus
 5 years ago
Best ResponseYou've already chosen the best response.0maybe i shouldn't have misused the v

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0completing 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
 5 years ago
Best ResponseYou've already chosen the best response.1...so partial fractions works too then I guess

UnkleRhaukus
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int{dv \over (2v+1)^2}=\int {dx \over x}\]\[{1 \over 4v+2}=ln(x) +c_1\]\[4v+2 = ln(1/x)+c_2\]

TuringTest
 5 years ago
Best ResponseYou've already chosen the best response.1haha, oh yeah.... I was overcomplicating it

UnkleRhaukus
 5 years ago
Best ResponseYou've already chosen the best response.0\[y=x/4(c_3lnx)\]

UnkleRhaukus
 5 years ago
Best ResponseYou've already chosen the best response.0oh wait, the answer in my book has arctans and lns

UnkleRhaukus
 5 years ago
Best ResponseYou've already chosen the best response.0i must have put a negative in the wrong stop somewhere

UnkleRhaukus
 5 years ago
Best ResponseYou've already chosen the best response.0the 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
 5 years ago
Best ResponseYou've already chosen the best response.0someone put me on the right track,

UnkleRhaukus
 5 years ago
Best ResponseYou've already chosen the best response.0where did i go wrong
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