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 April 6th, 2017, 07:34 PM #1 Member   Joined: Apr 2017 From: India Posts: 48 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: (dx-dy)/(y-z) and (dy+2dz)/(7(y+2z)) How each of these fractions can be equal to these fractions? Is there the use of componendo-dividendo 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:$${(dx-dy)\over(y-z)}\; \text{and}\; {(dy+2dz)\over7(y+2z)}$$Must be$${(dx-dz)\over(y-z)}\; \text{and}\; {(dy+2dz)\over7(y+2z)}$$and these are equal to any of the initial fractions. e.g.$${dx\over(-1)}= {(dx-dz)\over(y-z)}$$
April 7th, 2017, 09:43 PM   #3
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Quote:
 Originally Posted by deesuwalka Must be$${(dx-dz)\over(y-z)}\; \text{and}\; {(dy+2dz)\over7(y+2z)}$$and these are equal to any of the initial fractions. e.g.$${dx\over(-1)}= {(dx-dz)\over(y-z)}$$

$${(d\color{red}y-dz)\over(y-z)}\; \text{and}\; {(dy+2dz)\over7(y+2z)}$$and these are equal to any of the initial fractions. e.g.$${dx\over(-1)}= {(d\color{red}y-dz)\over(y-z)}$$

simultaneous differential equation

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