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Solve $$\frac{dx}{y+z}=\frac{dy}{z+x}=\frac{dz}{x+y}.$$ Solve the similtaneous equation by using method of multipliers. How can we choose these multipliers? Is there any specific method to know these multipliers?

Abelois
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raju
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2 Answers2

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Try to find triplets (multipliers) $l,m,n$ such that in,

$$\frac{dx}{y+z}=\frac{dy}{z+x}=\dfrac{dz}{x+y}=\dfrac{l dx+m dy+n dz}{l(y+z)+m(z+x)+n (x+y)}$$

make the denominator vanish

Set $l=y-z,m=z-x,n=x-y$

Checking for the denominator to vanish: $(y-z)(y+z)+(z-x)(z+x)+(x-y)(x+y)=y^2-z^2+z^2-x^2+x^2-y^2=0$

$$\frac{dx}{y+z}=\frac{dy}{z+x}=\dfrac{dz}{x+y}=\dfrac{(y-z)dx+(z-x)dy+(x-y)dz}{0}$$

We have to solve $(y-z)dx+(z-x)dy+(x-y)dz=0$ Rearranging,

$(y-z)dx+(z-y+y-x)dy+(x-y)dz=0$

$(y-z)dx-(y-z)dy-(x-y)dy+(x-y)dz=0$

$(y-z)(dx-dy)-(x-y)(dy-dz)=0$

$\dfrac{(dx-dy)(y-z)-(x-y)(dy-dz)}{(y-z)^2}=0$

Finally

$\dfrac{x-y}{y-z}=c_1$

Choosing another triplet in a similar fashion, you can get the second equation.

The specifity of the method is the posing of the quotient. With respecto to something more systematic, I don't think there is an algorithm to find them.

Rafa Budría
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  • Thanks. I already know that method. But in some problems it is too complicated to me to choose l,m,n for example xdx/(z^2-2yz-y^2)=dy/y+z=dz/y-z. To this how we choose multifliers? – raju Apr 10 '17 at 01:48
  • Yes, sometimes it's difficult, as everything in relation to differential equations. In this topic there are methods prescribed for some situations but without guarantee. I dealed a good amoung of time with your equation to get the multipliers because it seemed you need to apply this method, but in this case is easier manipulate the proportions directly. The equation you've shown in the comment is relatively easy to solve with multipliers. – Rafa Budría Apr 10 '17 at 07:18
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$$\frac{dx}{y+z}=\frac{dy}{z+x}=\frac{dz}{x+y}.$$ $$\frac{dx}{y+z}=\frac{dy}{z+x}=\frac{dx-dy}{y-x}.$$ $$\frac{dy}{z+x}=\frac{dz}{x+y}=\frac{dy-dz}{z-y}.$$ Therefore $$\frac{dx-dy}{y-x}=\frac{dy-dz}{z-y}.$$ $$\frac{d(x-y)}{x-y}=\frac{d(y-z)}{y-z}.$$ $$\ln{(x-y)}=\ln{(y-z)}+K$$ $$x=y+K({y-z})$$

user577215664
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