UnkleRhaukus
  • UnkleRhaukus
(3x-2y)dy+(x+y-5)dx=0
Differential Equations
jamiebookeater
  • jamiebookeater
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UnkleRhaukus
  • UnkleRhaukus
\[\newcommand \p \newcommand \p \dd [1] { \,\mathrm d#1 } % infinitesimal \p \de [2] { \frac{ \mathrm d #1}{\mathrm d#2} } % first order derivative \small\begin{align} (3x-2y)\dd y+(x+y-5)\dd x &=0 \\ \de yx &=\frac{5-x-y}{3x-2y} \\ \\ \text{let } x=X+p \\ y=Y+q \\ \de yx=\de YX \\ \\ \de YX &=\frac{5-(X+p)-(Y+q)}{3(X+p)-2(Y+q)} \\ \de YX &=\frac{-X-Y+(5-p-q)}{3X-2Y+(3p-2q)} \\ \\ 5-p-q=0 &\to q=5-p \\ 3p-2q=0 &\to 3p-2(5-p)=0 \to 5p-10=0 &\to p=2 \\ 5-2-q=0 &\to 3-q=0 \to q=3 \\ \\ \de YX &=\frac{-X-Y}{3X-2Y} \\ \de YX &=\frac{-1-Y/X}{3-2Y/X} \\ \\ \text{let }Y/X=V \\ Y=VX \\ Y'=V'X+V \\ \\ V'X+V &=\frac{-1-V}{3-2V} \\ V'X &=\frac{-1-V}{3-2V}-V\frac{3-2V}{3-2V} \\ V'X &=\frac{2V^2-4V-1}{3-2V} \\ \frac{3-2V}{2V^2-4V-1}\dd V &=\frac{\dd X}X \\ \int\frac{1}{2V^2-4V-1}\dd V +\int\frac{2-2V}{2V^2-4V-1}\dd V &=\int\frac{\dd X}X \\ \int\frac{1}{(\sqrt2V-\sqrt2)^2-3}\dd V -\frac12\int\frac{4V-4}{2V^2-4V-1}\dd V &=\int\frac{\dd X}X \\ -\frac13\int\frac{1}{1-\left(\sqrt{\tfrac23}(V-1)\right)^2}\dd V -\tfrac12\ln|2V^2-4V-1| &=\ln|X|+c \\ -\tfrac13\sqrt{\tfrac32}\tanh^{-1}\left(\sqrt{\tfrac23}(V-1)\right) -\tfrac12\ln|2V^2-4V-1| &=\ln|X|+c \\ -\tfrac1{\sqrt6}\tanh^{-1}\left(\sqrt{\tfrac23}\big(\tfrac{y-3}{x-2}-1\big)\right) -\tfrac12\ln\left|2(\tfrac{y-3}{x-2})^2-4(\tfrac{y-3}{x-2})-1\right| &=\ln|x-2|+c \end{align} \]
Psymon
  • Psymon
Checking answer? O.o
UnkleRhaukus
  • UnkleRhaukus
is there a more elegant method ?

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zzr0ck3r
  • zzr0ck3r
lol
anonymous
  • anonymous
subtract, integrate and then solve for y
Psymon
  • Psymon
Probably 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.
UnkleRhaukus
  • UnkleRhaukus
'x=dx/dx=1
anonymous
  • anonymous
No, there isn not. :( in my opinion this is one of those problems best left to a computer. :) http://www.wolframalpha.com/input/?i=(3x-2y)dy%2B(x%2By-5)dx%3D0&t=crmtb01
UnkleRhaukus
  • UnkleRhaukus
It's a question from a second year university differential equations course tutorial week 1
Psymon
  • Psymon
Seems pretty standard, really. Wish I could say Ive seen one of those y = vx dy = v'x + vx' subs that was pretty.
anonymous
  • anonymous
it looks like you drew an ostrich

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