Question

Find the general solution of
(y + 2xz)Zx - (x+2yZ)Zy = 1/2(x^2-y^2) X E R ; y>0

Answer 1

Answer on Question #78908 – Math – Differential Equations

Question

Find the general solution of

$$(y + 2xz)z_x - (x + 2yz)z_y = \frac{1}{2}(x^2 - y^2), \quad x \in \mathbb{R}; y > 0$$

Solution

$$\frac{dx}{y + 2xz} = \frac{-dy}{x + 2yz} = \frac{dz}{\frac{1}{2}(x^2 - y^2)}$$ $$\frac{dx - dy}{y + x + 2z(x + y)} = \frac{dz}{\frac{1}{2}(x^2 - y^2)}$$ $$\frac{d(x - y)}{(x + y)(2z + 1)} = \frac{2dz}{x^2 - y^2}$$ $$\frac{d(x - y)}{2z + 1} = \frac{2dz}{x - y}$$ $$\int (x - y)d(x - y) = 2\int (2z + 1)dz$$

Answer: $\frac{(x - y)^2}{2} = 2(z^2 + z) + C$.

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