Finding the Orthogonal Trajectory for a Family of Curves

[Mechdude]

Homework Statement

i want to get the orthogonal trajectory of the curves of this family

$$x^2 + y^2=cx$$

Homework Equations

answer is given as : $$y^2 + x^2=cy$$

The Attempt at a Solution

$$2x + 2yy' = \frac {x^2 +y^2} {x}$$ then $$y' = \frac{y} {2x} - \frac{x}{y}$$
let v=y/x ;
$$x\frac{dv}{dx} =\frac {-1}{2v}$$
thus:
$$- v^2 = \ln |x|$$ or
$$xe^{\frac {y^2}{x^2} } = c$$
which is far from the given answer . Got problem from odinary differential equations by rahman volume 1, 1994,

 

[Billy Bob]

then $$y' = \frac{y} {2x} - \frac{x}{y}$$

Should be $$y' = \frac{y} {2x} - \frac{x}{2y}$$

let v=y/x ;
$$x\frac{dv}{dx} =\frac {-1}{2v}$$

I didn't get anything close to that, even using the incorrect y'.

$$y' = \frac{y} {2x} - \frac{x}{2y} =\frac{y^2-x^2}{2xy}$$

so the orthogonal trajectory satisfies

$$y' = -\frac{2xy}{y^2-x^2}$$

Then use v=y/x as you suggested

 

[Mechdude]

Should be $$y' = \frac{y} {2x} - \frac{x}{2y}$$



I didn't get anything close to that, even using the incorrect y'.

$$y' = \frac{y} {2x} - \frac{x}{2y} =\frac{y^2-x^2}{2xy}$$

so the orthogonal trajectory satisfies

$$y' = -\frac{2xy}{y^2-x^2}$$

Then use v=y/x as you suggested

thanks for the corrections , i made errors in my working , so following from where u left i get stuck here; $$\frac{(v^2 -1)dv} {-v(v^2 + 1)} = \frac {dx}{x}$$

 

[Mechdude]

Thanks was able to integrate where billy bob left of using integrating factor$$\frac {1}{y^2}$$
for anyone who gets stranded

 

[Billy Bob]

Here's how I got past that step you mentioned:

$$-\frac{v^2-1}{v^3+v}=\frac{1}{v}-\frac{2v}{v^2+1}$$

Glad it worked out for you.