Question

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

Answer 1

(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

Answer 2

i tried that but i was confuse on what to do next

Answer 3

(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

Answer 4

What exactly are we solving for? Just curious.

Answer 5

we want to find the general solution y(x)
that satisfies that differential equation,

Answer 6

So that correlation \(u=\dfrac{y}{x}\) is the key in finding the general solution \(y(x)\).

Answer 7

correct

Answer 8

\[\frac{1+u}{1-u}=\frac{du}{dx}(x)+u\]\[\frac{1+\dfrac{y}{x}}{1-\dfrac{y}{x}}=\frac{du}{dx}(x)+\dfrac{y}{x}\]\[\frac{\dfrac{x+y}{x}}{\dfrac{x-y}{x}} = \frac{du}{x}(x) +\frac{y}{x}\]

Answer 9

\[\frac{x+y}{x-y}=\frac{du}{dx}(x)+\frac{y}{x}\]YEah I made a typo up there lol

Answer 10

(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

Answer 11

integrate both sides

Answer 12

\[\frac{x+y}{x-y}-\frac{y}{x}=\frac{du}{dx}(x)\]\[\frac{x(x+y)-y(x-y)}{x(x-y)}=\frac{du}{dx}(x)\]Hmmmmm oh.

Answer 13

Am I even on the right track? lol

Answer 14

Idk what to do with the RHS :\

Answer 15

ln|x| = arctan(u) - 1/2 ln(1 + u^2) + c

Answer 16

$$ \large {
(x + y) dx - (x-y) dy= 0
\\ (x+y)dx = (x-y) dy
\\xdx +ydx = xdy -ydy
\\ \text{divide through by x }
\\ dx + \frac y x \cdot dx = dy - \frac y x \cdot dy
\\(1 + \frac y x) \cdot dx = (1 - \frac y x) \cdot dy
\\~\\ \mathrm{Let~ u = \frac y x}
\\(1 + u) dx = (1- u ) dy
\\ \frac{ 1 + u }{ 1 - u} = \frac {dy}{dx }
\\ u = \frac y x \implies y = u \cdot x
\\\frac{dy}{dx} = \frac{du}{dx} \cdot x + u \cdot 1
\\ \mathrm{ substitute }
\\\frac{1 + u}{1 - u} =\frac{ du}{dx} \cdot x + u
\\ \text{this is separable D.E.}
\\ \ \\ \frac{1 + u}{1 - u} - u = \frac{ du}{dx} \cdot x
\\~\\ \frac{1 + u }{1-u} - u ~\frac{1 - u}{1 - u} = \frac{ du}{dx} \cdot x
\\~\\ \frac{1 + u - u + u^2}{1 - u} = \frac{ du}{dx}\cdot x
\\~\\ \frac{ 1 + u^2}{1-u} = \frac {du}{dx} \cdot x
\\~\\ \frac 1 x \cdot dx \cdot \frac{( 1 + u^2) }{ (1-u)} = du
\\ \frac 1 x \cdot dx = \frac{1-u}{1+u^2} ~du
\\~ \\ ∫ \frac 1 x \cdot dx =∫\frac {1-u}{ 1+u^2 }du
\\ ~\\ \ln |x | = ∫ \frac 1 {1 + u^2} - ∫ \frac u {1 + u^2}
\\ \ln |x | = \arctan(u) - \frac 1 2 ∫ \frac {2u} {1 + u^2}
\\ \ln |x | = \arctan(u) - \frac 1 2 \ln ( 1 + u^2) + C
\\ \text { substitute back}
\\ \ln |x | = \arctan(\frac y x ) - \frac 1 2 \ln \left( 1 + \frac {y^2}{x^2} \right) + C
}$$

Answer 17

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 :)