Question

find the integral curve of
a dx / (b-c)yz = b dy / (c-a)zx = c dz / (a-b)xy​

Answer 1

Answer:

Using the first two equality you can find the second one. Indeed,

adx/(b−c)yz = bdy/(c−a)zx ⇒ adx/(b−c)y = bdy/(c−a)x

⇒ axdx/(b−c) = bydy/(c−a) (a separable ODE)

⇒ ax2/(b−c)−by2 / (c−a) = c2

= v(x,y,z)

Step-by-step explanation:

Answer 2

Given:

adx / (b-c)yz = bdy / (c-a)zx = cdz / (a-b)xy​

To Find:

the integral curve of given equation

Explanation:

The equation is:

adx /(b−c)yz = bdy /(c−a)zx = cdz /(a−b)xy

We found one integral curve as such u=(ax²)/2 + (by²)/2 + (cz²)/2 = c, but we could not find the other curve.

Using the first two equality you can find the second one.

a dx /(b−c)yz = b dy /(c−a)zx

= a dx /(b−c)y = b dy(c−a)x

= axdx /(b−c) = bydy /(c−a)       (a separable ODE)

= ax²(b−c)−by²(c−a)

= c₂= v(x,y,z).

c₂ = v(x,y,z).

Answer - c₂ = v(x,y,z).