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anonymous
 one year ago
Find a oneparameter family of solutions to the isobaric equation
(x+y)dx(xy)dy=0
I tried to distribute but i got stuck srs
anonymous
 one year ago
Find a oneparameter family of solutions to the isobaric equation (x+y)dx(xy)dy=0 I tried to distribute but i got stuck srs

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0(x+y)dx(xy)dy=0. divide both sides by dx (x+y)  (xy)*dy/dx = 0 (x + y ) = (x  y) * dy/dx (x+y) / ( x  y) = dy/dx this doesnt' seem to help

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i tried that but i was confuse on what to do next

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0(x + y) dx  (xy) dy= 0 (x+y)dx = (xy) dy xdx +ydx = xdy ydy divide through by x dx + y/x*dx = dy + y/x*dy (1 + y/x) * dx = (1  y/x) *dy Let u = y/x ( 1 + u) dx = (1 u ) dy ( 1 + u ) / ( 1  u) = dy/dx Since u = y/x y = u*x dy/dx = du/dx * x + u*1 substitute ( 1 + u ) / ( 1  u) = du/dx * x + u*1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What exactly are we solving for? Just curious.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0we want to find the general solution y(x) that satisfies that differential equation,

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So that correlation \(u=\dfrac{y}{x}\) is the key in finding the general solution \(y(x)\).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{1+u}{1u}=\frac{du}{dx}(x)+u\]\[\frac{1+\dfrac{y}{x}}{1\dfrac{y}{x}}=\frac{du}{dx}(x)+\dfrac{y}{x}\]\[\frac{\dfrac{x+y}{x}}{\dfrac{xy}{x}} = \frac{du}{x}(x) +\frac{y}{x}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{x+y}{xy}=\frac{du}{dx}(x)+\frac{y}{x}\]YEah I made a typo up there lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0(x + y) dx  (xy) dy= 0 (x+y)dx = (xy) dy xdx +ydx = xdy ydy divide through by x dx + y/x*dx = dy + y/x*dy (1 + y/x) * dx = (1  y/x) *dy Let u = y/x ( 1 + u) dx = (1 u ) dy ( 1 + u ) / ( 1  u) = dy/dx Since u = y/x y = u*x dy/dx = du/dx * x + u*1 substitute (1 + u) / (1  u) = du/dx * x + u this is separable D.E. (1 + u) / (1  u)  u = du/dx * x (1 + u ) / (1u)  u (1u)/(1u) = du/dx * x (1 + u  u + u^2) / (1  u) = du/dx * x ( 1 + u^2) / (1u) = du/dx * x 1/x * dx * ( 1 + u^2) / (1u) = du 1/x * dx = (1u) / (1+u^2) du

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0integrate both sides

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{x+y}{xy}\frac{y}{x}=\frac{du}{dx}(x)\]\[\frac{x(x+y)y(xy)}{x(xy)}=\frac{du}{dx}(x)\]Hmmmmm oh.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Am I even on the right track? lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Idk what to do with the RHS :\

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0lnx = arctan(u)  1/2 ln(1 + u^2) + c

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0$$ \large { (x + y) dx  (xy) dy= 0 \\ (x+y)dx = (xy) dy \\xdx +ydx = xdy ydy \\ \text{divide through by x } \\ dx + \frac y x \cdot dx = dy  \frac y x \cdot dy \\(1 + \frac y x) \cdot dx = (1  \frac y x) \cdot dy \\~\\ \mathrm{Let~ u = \frac y x} \\(1 + u) dx = (1 u ) dy \\ \frac{ 1 + u }{ 1  u} = \frac {dy}{dx } \\ u = \frac y x \implies y = u \cdot x \\\frac{dy}{dx} = \frac{du}{dx} \cdot x + u \cdot 1 \\ \mathrm{ substitute } \\\frac{1 + u}{1  u} =\frac{ du}{dx} \cdot x + u \\ \text{this is separable D.E.} \\ \ \\ \frac{1 + u}{1  u}  u = \frac{ du}{dx} \cdot x \\~\\ \frac{1 + u }{1u}  u ~\frac{1  u}{1  u} = \frac{ du}{dx} \cdot x \\~\\ \frac{1 + u  u + u^2}{1  u} = \frac{ du}{dx}\cdot x \\~\\ \frac{ 1 + u^2}{1u} = \frac {du}{dx} \cdot x \\~\\ \frac 1 x \cdot dx \cdot \frac{( 1 + u^2) }{ (1u)} = du \\ \frac 1 x \cdot dx = \frac{1u}{1+u^2} ~du \\~ \\ ∫ \frac 1 x \cdot dx =∫\frac {1u}{ 1+u^2 }du \\ ~\\ \ln x  = ∫ \frac 1 {1 + u^2}  ∫ \frac u {1 + u^2} \\ \ln x  = \arctan(u)  \frac 1 2 ∫ \frac {2u} {1 + u^2} \\ \ln x  = \arctan(u)  \frac 1 2 \ln ( 1 + u^2) + C \\ \text { substitute back} \\ \ln x  = \arctan(\frac y x )  \frac 1 2 \ln \left( 1 + \frac {y^2}{x^2} \right) + C }$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ohhh, I was up to like line 8 starting from the bottom moving up. I dind't know what to do with the \(\dfrac{du}{dx}(x)\) ! This makes more sense, and also makes me happy :)
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