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April 6th, 2017, 07:34 PM   #1
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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.
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April 6th, 2017, 09:25 PM   #2
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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.$${dx\over(-1)}= {(dx-dz)\over(y-z)}$$
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April 7th, 2017, 09:43 PM   #3
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Quote:
Originally Posted by deesuwalka View Post
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.$${dx\over(-1)}= {(d\color{red}y-dz)\over(y-z)}$$
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