anonymous one year ago 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

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1. 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

2. anonymous

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

3. 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

4. Jhannybean

What exactly are we solving for? Just curious.

5. anonymous

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

6. Jhannybean

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

7. anonymous

correct

8. 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}$

9. Jhannybean

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

10. 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

11. anonymous

integrate both sides

12. 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.

13. Jhannybean

Am I even on the right track? lol

14. Jhannybean

Idk what to do with the RHS :\

15. anonymous

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

16. 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 } 17. 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 :)