Question

Solve the equation :

(7y-3x+3)dy + (3y-7x+7)dx =0

Answer 1

(7y-3x+3)dy + (3y-7x+7)dx =0

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

$\frac{dy}{dx}=\frac{3y-7x+7}{3x-7y-3}\\ now,let ~x=x'+h~and~y=y'+k\\ \frac{dy'}{dx'}=\frac{3y'-7x'+3k-7h+7}{3x'-7y'+3h-7k-3}\\ putting~h=1~and~k=0,we~get\\ \frac{dy'}{dx'}=\frac{3y'-7x'}{3x'-7y'}\\ now,let~ y'=vx'~so~that\\ \frac{dy'}{dx'}=v+x'\frac{dv}{dx'}\\ we~have\\ v+x'\frac{dv}{dx'}=\frac{3vx'-7x'}{3x'-7vx'}\\ x'\frac{dv}{dx'}=\frac{3v-7}{3-7v}-v\\ x'\frac{dv}{dx'}=\frac{3v-7-3v+7v^2}{3-7v}\\ x'\frac{dv}{dx'}=\frac{7v^2-7}{3-7v}\\ \frac{dx'}{x'}=\frac{3-7v}{7v^2-7}dv\\ integrating~both~sides\\ \int \frac{dx'}{x'}=\int \frac{3}{7v^2-7}dv-\int \frac{7v}{7v^2-7}dv\\ ln|x'|+c= \frac{3}{14}ln \frac{|v-1|}{|v+1|}-\frac{1}{2}ln|{v^2-1}|\\ ln|x'|+c= \frac{3}{14}ln \frac{|y'-x'|}{|y'+x'|}-\frac{1}{2} ln|\frac{(y')^2-(x')^2}{(x')^2}|\\ Recall~v=\frac{y'}{x'}\\ ln|x'|+c= \frac{3}{14}ln {|y'-x'|}- \frac{3}{14}{|y'+x'|}-\frac{1}{2} ln|{y'-x'}|+\frac{1}{2} ln|{y'+x'}+\frac{1}{2} ln|({x'})^2|\\ ln|x'|+c= -\frac{2}{7}ln {|y'-x'|}-\frac{5}{7}ln {|y'+x'|}+\frac{1}{2} ln|({x'})^2|\\ 2ln|y'-x'|+5ln|y'+x'|=C\\ 2ln|y-x+1|+5ln|y+x-1|=C\\ (Since~y'=y~and~x'=(x-1))$