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.