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UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1\[\newcommand \p \newcommand \p \dd [1] { \,\mathrm d#1 } % infinitesimal \p \de [2] { \frac{ \mathrm d #1}{\mathrm d#2} } % first order derivative \small\begin{align} (3x2y)\dd y+(x+y5)\dd x &=0 \\ \de yx &=\frac{5xy}{3x2y} \\ \\ \text{let } x=X+p \\ y=Y+q \\ \de yx=\de YX \\ \\ \de YX &=\frac{5(X+p)(Y+q)}{3(X+p)2(Y+q)} \\ \de YX &=\frac{XY+(5pq)}{3X2Y+(3p2q)} \\ \\ 5pq=0 &\to q=5p \\ 3p2q=0 &\to 3p2(5p)=0 \to 5p10=0 &\to p=2 \\ 52q=0 &\to 3q=0 \to q=3 \\ \\ \de YX &=\frac{XY}{3X2Y} \\ \de YX &=\frac{1Y/X}{32Y/X} \\ \\ \text{let }Y/X=V \\ Y=VX \\ Y'=V'X+V \\ \\ V'X+V &=\frac{1V}{32V} \\ V'X &=\frac{1V}{32V}V\frac{32V}{32V} \\ V'X &=\frac{2V^24V1}{32V} \\ \frac{32V}{2V^24V1}\dd V &=\frac{\dd X}X \\ \int\frac{1}{2V^24V1}\dd V +\int\frac{22V}{2V^24V1}\dd V &=\int\frac{\dd X}X \\ \int\frac{1}{(\sqrt2V\sqrt2)^23}\dd V \frac12\int\frac{4V4}{2V^24V1}\dd V &=\int\frac{\dd X}X \\ \frac13\int\frac{1}{1\left(\sqrt{\tfrac23}(V1)\right)^2}\dd V \tfrac12\ln2V^24V1 &=\lnX+c \\ \tfrac13\sqrt{\tfrac32}\tanh^{1}\left(\sqrt{\tfrac23}(V1)\right) \tfrac12\ln2V^24V1 &=\lnX+c \\ \tfrac1{\sqrt6}\tanh^{1}\left(\sqrt{\tfrac23}\big(\tfrac{y3}{x2}1\big)\right) \tfrac12\ln\left2(\tfrac{y3}{x2})^24(\tfrac{y3}{x2})1\right &=\lnx2+c \end{align} \]

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1is there a more elegant method ?

pgpilot326
 one year ago
Best ResponseYou've already chosen the best response.0subtract, integrate and then solve for y

Psymon
 one year ago
Best ResponseYou've already chosen the best response.0Probably not. I would have done basically the same thing except I might have done it right off the bat. y = vx and dy = v'x + vx' and went from there.

agentc0re
 one year ago
Best ResponseYou've already chosen the best response.0No, there isn not. :( in my opinion this is one of those problems best left to a computer. :) http://www.wolframalpha.com/input/?i=(3x2y)dy%2B(x%2By5)dx%3D0&t=crmtb01

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1It's a question from a second year university differential equations course tutorial week 1

Psymon
 one year ago
Best ResponseYou've already chosen the best response.0Seems pretty standard, really. Wish I could say Ive seen one of those y = vx dy = v'x + vx' subs that was pretty.

glewis15
 one year ago
Best ResponseYou've already chosen the best response.0it looks like you drew an ostrich
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