Answer to Question #288879 in Differential Equations for Pankaj

Question

Solve the equation :

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

Answer 1

Solution:

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

$\Rightarrow\dfrac{dy}{dx}=\dfrac{3y-7x+7}{3x-7y-3}\\ now,let ~x=x'+h~and~y=y'+k\\ \Rightarrow \dfrac{dy'}{dx'}=\dfrac{3y'-7x'+3k-7h+7}{3x'-7y'+3h-7k-3}\\ putting~h=1~and~k=0,we~get\\ \Rightarrow\dfrac{dy'}{dx'}=\dfrac{3y'-7x'}{3x'-7y'}\\ now,let~ y'=vx'~so~that\\ \Rightarrow\dfrac{dy'}{dx'}=v+x'\dfrac{dv}{dx'}\\ we~have\\ \Rightarrow v+x'\dfrac{dv}{dx'}=\dfrac{3vx'-7x'}{3x'-7vx'}\\ \Rightarrow x'\dfrac{dv}{dx'}=\dfrac{3v-7}{3-7v}-v\\ \Rightarrow x'\dfrac{dv}{dx'}=\dfrac{3v-7-3v+7v^2}{3-7v}\\ \Rightarrow x'\dfrac{dv}{dx'}=\dfrac{7v^2-7}{3-7v}\\ \Rightarrow\dfrac{dx'}{x'}=\dfrac{3-7v}{7v^2-7}dv\\ integrating~both~sides\\ \Rightarrow\int \dfrac{dx'}{x'}=\int \dfrac{3}{7v^2-7}dv-\int \dfrac{7v}{7v^2-7}dv\\ \Rightarrow ln|x'|+c= \dfrac{3}{14}ln \dfrac{|v-1|}{|v+1|}-\dfrac{1}{2}ln|{v^2-1}|\\ \Rightarrow ln|x'|+c= \dfrac{3}{14}ln \dfrac{|y'-x'|}{|y'+x'|}-\dfrac{1}{2} ln|\dfrac{(y')^2-(x')^2}{(x')^2}|\\ Recall~v=\dfrac{y'}{x'}\\ \Rightarrow ln|x'|+c= \dfrac{3}{14}ln {|y'-x'|}- \dfrac{3}{14}{|y'+x'|}-\dfrac{1}{2} ln|{y'-x'}|+\dfrac{1}{2} ln|{y'+x'}+\dfrac{1}{2} ln|({x'})^2|\\ \Rightarrow ln|x'|+c= -\dfrac{2}{7}ln {|y'-x'|}-\dfrac{5}{7}ln {|y'+x'|}+\dfrac{1}{2} ln|({x'})^2|\\ \Rightarrow 2ln|y'-x'|+5ln|y'+x'|=C\\ \Rightarrow2ln|y-x+1|+5ln|y+x-1|=C\\ (Since~y'=y~and~x'=(x-1))$