Question

(y+z)p_(z+x)q=x_y

Answer 1

Answer on Question #82198 – Math – Differential Equations

Question

$$(y + z) p - (z + x) q = x - y$$

Solution

Compare with $Pp + Qq = R$

$$P = y + z, Q = -(z + x), R = x - y$$

Consider Lagrange’s auxiliary equations

$$\frac{dx}{P} = \frac{dy}{Q} = \frac{dz}{R}$$ $$\frac{dx}{y + z} = \frac{dy}{-(z + x)} = \frac{dz}{x - y} = \frac{dx + dy + dz}{y + z - z - x + x - y} = \frac{dx + dy + dz}{0}$$ $$dx + dy + dz = 0 \Rightarrow x + y + z = c_1$$ $$\frac{dx}{y + z} = \frac{dy}{-(z + x)} = \frac{dz}{x - y} = \frac{x dx + y dy - z dz}{x y + x z - y z - x y - x z + y z} = \frac{x dx + y dy - z dz}{0}$$ $$d(x^2 + y^2 - z^2) = 0 \Rightarrow x^2 + y^2 - z^2 = c_2$$

General solution is

$$F(x + y + z, x^2 + y^2 - z^2) = 0$$

Answer: $F(x + y + z, x^2 + y^2 - z^2) = 0.$

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