anonymous
  • anonymous
Find a one-parameter family of solutions to the isobaric equation (x+y)dx-(x-y)dy=0 I tried to distribute but i got stuck srs
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
(x+y)dx-(x-y)dy=0. divide both sides by dx (x+y) - (x-y)*dy/dx = 0 (x + y ) = (x - y) * dy/dx (x+y) / ( x - y) = dy/dx this doesnt' seem to help
anonymous
  • anonymous
i tried that but i was confuse on what to do next
anonymous
  • anonymous
(x + y) dx - (x-y) dy= 0 (x+y)dx = (x-y) 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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

Jhannybean
  • Jhannybean
What exactly are we solving for? Just curious.
anonymous
  • anonymous
we want to find the general solution y(x) that satisfies that differential equation,
Jhannybean
  • Jhannybean
So that correlation \(u=\dfrac{y}{x}\) is the key in finding the general solution \(y(x)\).
anonymous
  • anonymous
correct
Jhannybean
  • Jhannybean
\[\frac{1+u}{1-u}=\frac{du}{dx}(x)+u\]\[\frac{1+\dfrac{y}{x}}{1-\dfrac{y}{x}}=\frac{du}{dx}(x)+\dfrac{y}{x}\]\[\frac{\dfrac{x+y}{x}}{\dfrac{x-y}{x}} = \frac{du}{x}(x) +\frac{y}{x}\]
Jhannybean
  • Jhannybean
\[\frac{x+y}{x-y}=\frac{du}{dx}(x)+\frac{y}{x}\]YEah I made a typo up there lol
anonymous
  • anonymous
(x + y) dx - (x-y) dy= 0 (x+y)dx = (x-y) 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 ) / (1-u) - u (1-u)/(1-u) = du/dx * x (1 + u - u + u^2) / (1 - u) = du/dx * x ( 1 + u^2) / (1-u) = du/dx * x 1/x * dx * ( 1 + u^2) / (1-u) = du 1/x * dx = (1-u) / (1+u^2) du
anonymous
  • anonymous
integrate both sides
Jhannybean
  • Jhannybean
\[\frac{x+y}{x-y}-\frac{y}{x}=\frac{du}{dx}(x)\]\[\frac{x(x+y)-y(x-y)}{x(x-y)}=\frac{du}{dx}(x)\]Hmmmmm oh.
Jhannybean
  • Jhannybean
Am I even on the right track? lol
Jhannybean
  • Jhannybean
Idk what to do with the RHS :\
anonymous
  • anonymous
ln|x| = arctan(u) - 1/2 ln(1 + u^2) + c
anonymous
  • anonymous
$$ \large { (x + y) dx - (x-y) dy= 0 \\ (x+y)dx = (x-y) dy \\xdx +ydx = xdy -ydy \\ \text{divide through by x } \\ dx + \frac y x \cdot dx = dy - \frac y x \cdot dy \\(1 + \frac y x) \cdot dx = (1 - \frac y x) \cdot dy \\~\\ \mathrm{Let~ u = \frac y x} \\(1 + u) dx = (1- u ) dy \\ \frac{ 1 + u }{ 1 - u} = \frac {dy}{dx } \\ u = \frac y x \implies y = u \cdot x \\\frac{dy}{dx} = \frac{du}{dx} \cdot x + u \cdot 1 \\ \mathrm{ substitute } \\\frac{1 + u}{1 - u} =\frac{ du}{dx} \cdot x + u \\ \text{this is separable D.E.} \\ \ \\ \frac{1 + u}{1 - u} - u = \frac{ du}{dx} \cdot x \\~\\ \frac{1 + u }{1-u} - u ~\frac{1 - u}{1 - u} = \frac{ du}{dx} \cdot x \\~\\ \frac{1 + u - u + u^2}{1 - u} = \frac{ du}{dx}\cdot x \\~\\ \frac{ 1 + u^2}{1-u} = \frac {du}{dx} \cdot x \\~\\ \frac 1 x \cdot dx \cdot \frac{( 1 + u^2) }{ (1-u)} = du \\ \frac 1 x \cdot dx = \frac{1-u}{1+u^2} ~du \\~ \\ ∫ \frac 1 x \cdot dx =∫\frac {1-u}{ 1+u^2 }du \\ ~\\ \ln |x | = ∫ \frac 1 {1 + u^2} - ∫ \frac u {1 + u^2} \\ \ln |x | = \arctan(u) - \frac 1 2 ∫ \frac {2u} {1 + u^2} \\ \ln |x | = \arctan(u) - \frac 1 2 \ln ( 1 + u^2) + C \\ \text { substitute back} \\ \ln |x | = \arctan(\frac y x ) - \frac 1 2 \ln \left( 1 + \frac {y^2}{x^2} \right) + C }$$
Jhannybean
  • Jhannybean
Ohhh, 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 :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.