Question

(3x-2y)dy+(x+y-5)dx=0

Answer 1

\[\newcommand \p \newcommand
\p \dd [1] { \,\mathrm d#1 } % infinitesimal
\p \de [2] { \frac{ \mathrm d #1}{\mathrm d#2} } % first order derivative
\small\begin{align} (3x-2y)\dd y+(x+y-5)\dd x &=0 \\
\de yx &=\frac{5-x-y}{3x-2y} \\
\\
\text{let } x=X+p \\
y=Y+q \\
\de yx=\de YX \\
\\
\de YX &=\frac{5-(X+p)-(Y+q)}{3(X+p)-2(Y+q)} \\
\de YX &=\frac{-X-Y+(5-p-q)}{3X-2Y+(3p-2q)} \\
\\
5-p-q=0 &\to q=5-p \\
3p-2q=0 &\to 3p-2(5-p)=0 \to
5p-10=0 &\to p=2 \\
5-2-q=0 &\to 3-q=0 \to q=3 \\
\\
\de YX &=\frac{-X-Y}{3X-2Y} \\
\de YX &=\frac{-1-Y/X}{3-2Y/X} \\
\\
\text{let }Y/X=V \\
Y=VX \\
Y'=V'X+V \\
\\
V'X+V &=\frac{-1-V}{3-2V} \\
V'X &=\frac{-1-V}{3-2V}-V\frac{3-2V}{3-2V} \\
V'X &=\frac{2V^2-4V-1}{3-2V} \\
\frac{3-2V}{2V^2-4V-1}\dd V &=\frac{\dd X}X \\
\int\frac{1}{2V^2-4V-1}\dd V
+\int\frac{2-2V}{2V^2-4V-1}\dd V &=\int\frac{\dd X}X \\
\int\frac{1}{(\sqrt2V-\sqrt2)^2-3}\dd V
-\frac12\int\frac{4V-4}{2V^2-4V-1}\dd V &=\int\frac{\dd X}X \\
-\frac13\int\frac{1}{1-\left(\sqrt{\tfrac23}(V-1)\right)^2}\dd V
-\tfrac12\ln|2V^2-4V-1| &=\ln|X|+c \\
-\tfrac13\sqrt{\tfrac32}\tanh^{-1}\left(\sqrt{\tfrac23}(V-1)\right)
-\tfrac12\ln|2V^2-4V-1| &=\ln|X|+c \\
-\tfrac1{\sqrt6}\tanh^{-1}\left(\sqrt{\tfrac23}\big(\tfrac{y-3}{x-2}-1\big)\right)
-\tfrac12\ln\left|2(\tfrac{y-3}{x-2})^2-4(\tfrac{y-3}{x-2})-1\right|
&=\ln|x-2|+c
\end{align}
\]

Answer 2

Checking answer? O.o

Answer 3

is there a more elegant method ?

Answer 4

lol

Answer 5

subtract, integrate and then solve for y

Answer 6

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.

Answer 7

'x=dx/dx=1

Answer 8

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

Answer 9

It's a question from a second year university differential equations course tutorial week 1

Answer 10

Seems pretty standard, really. Wish I could say Ive seen one of those y = vx dy = v'x + vx' subs that was pretty.

Answer 11

it looks like you drew an ostrich