How will you solve the following non-homogenous differential equation?

Question

here's the question

(3x-7y-3)dy = (3y-7x+7)dx

i have problems in doing the integration part. so, a clear explanation of the problem will be very helpful.

Answer 1

(3x−7y−3)dy=(3y−7x+7)dx

or, (dy/dx)=(3y−7x+7)/(3x−7y−3)

Now, let x=x’+h and y=y’+k

Therefore, (dy’/dx’)

=(3y’−7x’+3k−7h+7)/(3x’−7y’+3h−7k−3)

Putting h=1 and k=0, we get,

(dy’/dx’)=(3y’−7x’)/(3x’−7y’)

Now, let y’=vx’ so that (dy’/dx’)=v+x’(dv/dx’)

We have, v+x’(dv/dx’)=(3vx’−7x’)/(3x’−7vx’)

i.e. x’(dv/dx’)=[(3v−7)/(3−7v)]−v

i.e. x’(dv/dx’)=(3v−7−3v+7v²)/(3−7v)

i.e. x’(dv/dx’)=(7v²−7)/(3−7v)

i.e. dx’/x’=[(3−7v)/(7v²−7)]dv

Integrating both sides,

∫(dx’/x’)=∫[3 dv/(7v²−7)]−∫[7v dv/(7v²−7)]

i.e. ∫(dx’/x’)=(3/7)∫[dv/(v²−1)]−∫[v dv/(v²−1)]

i.e. ln|x’|+c=(3/7)(1/2)ln|(v−1)/(v+1)]

−(1/2)∫[d(v²−1)/(v²−1)]

i.e. ln|x’|+c=(3/14)ln|(v−1)/(v+1)|

−(1/2)ln|v²−1|

i.e. ln|x’|+c=(3/14)ln|(y’−x’)/(y’+x’)|

−(1/2)ln|(y’²−x’²)/x’²|

[since v=y’/x’]

i.e. ln |x’|+c=(3/14)ln|y’−x’|−(3/14)ln|y’+x’|−

(1/2)ln|y’+x’|−(1/2)ln|y’−x’|

−(1/2)ln|x’²|

i.e. ln |x’|+c=−(2/7)ln|y’−x’|−(5/7)ln|y’+x’|−ln...

i.e. 2ln|y’−x’|+5ln|y’+x’|=C

i.e. 2ln|y−x+1|+5ln|y+x−1|=C

[since y’=y and x’=(x−1)]

The process is okay but please check if there are any silly mistakes of calculation (after all, I had to do it very quickly). I beg your pardon in advance.