Question

PDE

1.2xyp+(x^2+y^2)q=(x+y)z
2.y(x+z)p+(z^2-2xz-x^2q=y(x-z)


Plz fast

Answer 1

Answer on Question #82221 - Math - Differential Equations

Question

1.

$$2 \times y p + (x ^ {2} + y ^ {2}) q = (x + y) z$$

Solution

$$\frac {d x}{2 x y} = \frac {d y}{x ^ {2} + y ^ {2}} = \frac {d z}{x + y}$$ $$\frac {d x + d y}{2 x y + x ^ {2} + y ^ {2}} = \frac {d z}{x + y}$$ $$\frac {d (x + y)}{(x + y) ^ {2}} = \frac {d z}{x + y}$$ $$\frac {d (x + y)}{x + y} = d z$$ $$z = \ln (c (x + y))$$

Answer: $z = \ln \bigl (c(x + y)\bigr)$

Question

2.

$$y (x + z) p + (z ^ {2} - 2 x z - x ^ {2}) q = y (x - z)$$

Solution

$$\frac {d x}{y (x + z)} = \frac {d y}{z ^ {2} - 2 x z - x ^ {2}} = \frac {d z}{y (x - z)}$$ $$\frac {d x}{x + z} = \frac {d z}{x - z}$$

Substitute:

$$z = t x; z ^ {\prime} = t ^ {\prime} x + t$$ $$t ^ {\prime} x + t = \frac {x - t x}{x + t x}$$ $$(t'x + t)(x + tx) = x - tx$$ $$t'x^2 + t'tx^2 + tx + t^2x = x - tx$$ $$x(t + 1)t' = 1 - 2t - t^2$$ $$\int \frac{t + 1}{1 - 2t - t^2} dt = \int \frac{dx}{x}$$ $$\frac{t + 1}{1 - 2t - t^2} = -\frac{t + 1}{(t + 1 + \sqrt{2})(t + 1 - \sqrt{2})}$$ $$\frac{t + 1}{(t + 1 + \sqrt{2})(t + 1 - \sqrt{2})} = \frac{A}{t + 1 + \sqrt{2}} + \frac{B}{t + 1 - \sqrt{2}}$$ $$A(t + 1 - \sqrt{2}) + B(t + 1 + \sqrt{2}) = t + 1$$ $$A + B = 1$$ $$A(1 - \sqrt{2}) + B(1 + \sqrt{2}) = 1$$ $$A = B = 0.5$$ $$\begin{aligned} & \int \frac{t + 1}{1 - 2t - t^2} dt = -\frac{1}{2} \int \frac{dt}{t + 1 + \sqrt{2}} - \frac{1}{2} \int \frac{dt}{t + 1 - \sqrt{2}} = -\frac{1}{2} \ln |t + 1 + \sqrt{2}| - \frac{1}{2} \ln |t + 1 - \sqrt{2}| = \\ & \quad = \ln \left| \frac{1}{\sqrt{t^2 + 2t - 1}} \right| \\ & \quad \ln \left| \frac{1}{\sqrt{t^2 + 2t - 1}} \right| = \ln |cx| \\ & \quad \frac{1}{\sqrt{t^2 + 2t - 1}} = cx \\ & \quad \frac{1}{\sqrt{\left(\frac{z}{x}\right)^2 + 2\frac{z}{x} - 1}} = cx \\ & \quad \frac{x}{\sqrt{z^2 + 2zx - x^2}} = cx \end{aligned}$$ $$z^{2} + 2zx - x^{2} = c_{1}$$

Now, choosing $(x,y,z)$ as multipliers, we get:

$$xdx + ydy + zdz = 0$$ $$x^{2} + y^{2} + z^{2} = c_{2}$$ $$\phi(z^{2} + 2zx - x^{2}, x^{2} + y^{2} + z^{2}) = 0$$

Answer: $\phi(z^{2} + 2zx - x^{2}, x^{2} + y^{2} + z^{2}) = 0$.

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