Question

x2(y-z) p+y2(z-x) q=z2(x-y) ​

Answer 1

Answer:

Step-by-step explanation:The Lagrange's Auxiliary Equations corresponding to the given equations are

dx/{x² (y - z)} = dy/{y² (z - x)} = dz/{z² (x - y)} ..... (1)

Using 1/x, 1/y, 1/z as multipliers for first, second and third ratio of (1), we write

each ratio of (1)

= (dx/x + dy/y + dz/z)/{x (y - z) + y (z - x) + z (x - y)}

= (dx/x + dy/y + dz/z)/(xy - zx + yz - xy + zx - yz)

= (dx/x + dy/y + dz/z)/0

∴ dx/x + dy/y + dz/z = 0

On integration,

logx + logy + logz = log c₁

or, xyz = c₁ ..... (2)

Again we choose 1/x², 1/y², 1/z² as multipliers of ratios of (1) and thus we get

each ratio of (1)

= (dx/x² + dy/y² + dz/z²)/(y - z + z - x + x - y)

= (dx/x² + dy/y² + dz/z²)/0

∴ dx/x² + dy/y² + dz/z² = 0

On integration,

- 1/x - 1/y - 1/z = - c₂

or, 1/x + 1/y + 1/z = c₂ ..... (3)

Hence, from (2) and (3), we obtain the general solution of the given equation as

φ(xyz, 1/x + 1/y + 1/z) = 0,

where φ is an arbitrary function.