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April 6th, 2017, 07:34 PM  #1 
Member Joined: Apr 2017 From: India Posts: 34 Thanks: 0  Simultaneous differential equation
While going through the concept of simultaneous differential equation, I came across following problem: Find the Integral curves of: dx/(1)= dy/(3y+4z)= dz/(2y+5z) The solution says: Each of these fractions of the given system of equations is equal to: (dxdy)/(yz) and (dy+2dz)/(7(y+2z)) How each of these fractions can be equal to these fractions? Is there the use of componendodividendo concept? If yes, how we deduced that? Last edited by skipjack; April 7th, 2017 at 11:17 PM. 
April 6th, 2017, 09:25 PM  #2 
Member Joined: Sep 2016 From: India Posts: 88 Thanks: 30 
The original equations form three proportions. Proportions fulfill this identity:$$\frac{a}{b}=\frac{c}{d}\implies\frac{a+ c}{b+d}=\frac{a}{b}\\ad=bc\implies ab+ad=ab+bc\implies a(b+d)=b(a+c)\implies\frac{a}{b}=\frac{a+c}{b+d}$$ So, this is wrong:$${(dxdy)\over(yz)}\; \text{and}\; {(dy+2dz)\over7(y+2z)}$$Must be$${(dxdz)\over(yz)}\; \text{and}\; {(dy+2dz)\over7(y+2z)}$$and these are equal to any of the initial fractions. e.g.$${dx\over(1)}= {(dxdz)\over(yz)}$$

April 7th, 2017, 09:43 PM  #3  
Member Joined: Sep 2016 From: India Posts: 88 Thanks: 30  Quote:
$${(d\color{red}ydz)\over(yz)}\; \text{and}\; {(dy+2dz)\over7(y+2z)}$$and these are equal to any of the initial fractions. e.g.$${dx\over(1)}= {(d\color{red}ydz)\over(yz)}$$  

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