Aspire Faculty ID #17134 · Topic: AMU MCA 2020 · Just now
AMU MCA 2020

The general solution of $(y - z)p + (z - x)q = x - y$ is

Solution

Given Lagrange’s linear PDE $(y - z)p + (z - x)q = x - y$ Auxiliary equations: $\frac{dx}{y - z} = \frac{dy}{z - x} = \frac{dz}{x - y}$ First integral: Add numerators and denominators: $(y - z) + (z - x) + (x - y) = 0$ ⇒ $dx + dy + dz = 0$ ⇒ $x + y + z = c_1$ Second integral: Multiply by $x, y, z$ respectively and add: $x(y - z) + y(z - x) + z(x - y) = 0$ ⇒ $d(x^2 + y^2 + z^2) = 0$ ⇒ $x^2 + y^2 + z^2 = c_2$ Hence general solution is $\phi(x + y + z,; x^2 + y^2 + z^2) = 0$
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