Question

The general solution of \(\frac{a \ dx}{(b-c)yz} = \frac{b \ dy}{(c - a)zx} = \frac{c \ dz}{(a - b)xy}\) will be _____, where c₁ and c₂ are arbitary constants.
1. ax² + by² - cz^₂ = c₁, a²x² - b²y² + c²z² = c₂
2. ax + by + cz = c₁, ax² = by² - by² - cz² = c₂
3. ax² + by² + cz² = c₁, a²x² + b²y² + c²z² = c₂
4. ax - by + cz = c₁, ax² + by² - cz² = c₂

Answer 1

Correct Answer - Option 3 : ax² + by² + cz² = c₁, a²x² + b²y² + c²z² = c₂

Explanation:

Using the Lagrange method of multipliers the differential equation can be solved.

Let:

\(\frac{{adx}}{{\left( {b - c} \right)yz}} = \frac{{bdy}}{{\left( {c - a} \right)zx}} = \frac{{cdz}}{{\left( {a - b} \right)xy}} = k\)

\( \Rightarrow \frac{{axdx}}{{\left( {b - c} \right)xyz}} = \frac{{bydy}}{{\left( {c - a} \right)xyz}} = \frac{{czdz}}{{\left( {a - b} \right)xyz}}\)

\( = \frac{{axdx + bydy + czdz}}{{\left( {b - c} \right)xyz + \left( {c - a} \right)xyz + \left( {a - b} \right)xyz}}\)

\(\frac{{axdx + bydy + czdz}}{0} = k\)

axdx + bydy + czdz = 0

Integrating, we get:

\(\frac{{a{x^2}}}{2} + \frac{{b{y^2}}}{2} + \frac{{c{z^2}}}{2} = {c_1}\)

∴ ax² + by² + cz² = 2c₁ = c = constant

Again,

\(\frac{{adx}}{{\left( {b - c} \right)yz}} = \frac{{bdy}}{{\left( {c - a} \right)zx}} = \frac{{cdz}}{{\left( {a - b} \right)xy}} = k,\) say

\( \Rightarrow \frac{{a.\left( {ax} \right)dx}}{{a\left( {b - c} \right)xyz}} = \frac{{b.\left( {by} \right)dy}}{{b\left( {c - a} \right)xyz}} = \frac{{c\left( {cz} \right)dz}}{{c\left( {a - b} \right)xyz}}\)

\(= \frac{{{a^2}xdx + {b^2}ydy + {c^2}zdz}}{{a\left( {b - c} \right)xyz + b\left( {c - a} \right)xyz + c\left( {a - b} \right)xyz}}\)

= k

\(\frac{{{a^2}xdx + {b^2}ydy + {c^2}zdz}}{0} = k\)

⇒ a²xdx + b²ydy + c²zdz = 0

Integrating, we get →

\(\frac{{{a^2}{x^2}}}{2} + \frac{{{b^2}{y^2}}}{2} + \frac{{{c^2}{z^2}}}{2} = {c_2}\)

a²x² + b²y² + c²z² = 2c₂ = c₃ (constant)