Solve partial differential equations x^2 (y-z)p+y^2 (z-x)q=z^2 (x-y)
Answers
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.