Question

Solve the differential equation (1) x^2p + y^2q = (x+y)z. (2) p^1/2 -q^1/2 +3x =0

Answer 1

Answer on Question #66338 – Math – Differential Equations

Question

Solve the differential equation

1)

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

Solution

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

Equation of characteristics:

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

Then:

$$\int \frac {d x}{x ^ {2}} = \int \frac {d y}{y ^ {2}} - \frac {1}{x} = - \frac {1}{y} + C _ {1}$$ $$C _ {1} = \frac {x - y}{x y}$$ $$y = \frac {x}{C _ {1} x + 1}$$ $$\frac {d x}{x ^ {2}} = \frac {d z}{(x + y) z} = \frac {d z}{\left(x + \frac {x}{C _ {1} x + 1}\right) z}$$ $$\frac {C _ {1} x ^ {2} + 2 x}{x ^ {2} (C _ {1} x + 1)} d x = \frac {C _ {1} x + 2}{x (C _ {1} x + 1)} = \frac {d z}{z}$$ $$\int \frac {C _ {1} x + 2}{x (C _ {1} x + 1)} d x = \int \frac {d z}{z}$$ $$\frac {C _ {1} x + 2}{x \left(C _ {1} x + 1\right)} = \frac {1}{x} + \frac {1}{x \left(C _ {1} x + 1\right)}$$ $$\frac {1}{x \left(C _ {1} x + 1\right)} = \frac {A}{x} + \frac {B}{C _ {1} x + 1}$$ $$A \left(C _ {1} x + 1\right) + B x = 1$$ $$A = 1; B = - C _ {1}$$ $$\frac {1}{x \left(C _ {1} x + 1\right)} = \frac {1}{x} - \frac {C _ {1}}{C _ {1} x + 1}$$ $$\int \left(\frac {2}{x} - \frac {C _ {1}}{C _ {1} x + 1}\right) d x = \int \frac {d z}{z}$$ $$2 \ln | x | - \ln | C _ {1} x + 1 | = \ln | z | + \ln C _ {2}$$ $$C _ {2} = \frac {x ^ {2}}{z \left(C _ {1} x + 1\right)} = \frac {x ^ {2}}{z \left(\frac {x - y}{x y} x + 1\right)} = \frac {y x ^ {2}}{z x} = \frac {x y}{z}$$

Answer: $F\left(\frac{x - y}{xy}, \frac{xy}{z}\right) = 0$

Question

2)

$$p ^ {1 / 2} - q ^ {1 / 2} + 3 x = 0$$

Solution

$$\left(\frac {\partial z}{\partial x}\right) ^ {1 / 2} - \left(\frac {\partial z}{\partial y}\right) ^ {1 / 2} + 3 x = 0$$

Equations of characteristics:

$$\dot {x} = \frac {d x}{d \tau} = \frac {1}{2 \sqrt {p}}; \dot {y} = \frac {d y}{d \tau} = - \frac {1}{2 \sqrt {q}}; \dot {p} = \frac {d p}{d \tau} = - 3; \dot {q} = \frac {d q}{d \tau} = 0; \dot {z} = \frac {d z}{d \tau} = \frac {\sqrt {p} - \sqrt {q}}{2}$$

Then:

$$p = -3 \int d\tau = -3\tau + C_1$$ $$q = C_2$$ $$y = -\int \frac{d\tau}{2\sqrt{q}} = -\frac{\tau}{2\sqrt{q}} + C_3 = -\frac{\tau}{2\sqrt{C_2}} + C_3$$ $$x = \int \frac{d\tau}{2\sqrt{p}} = \int \frac{d\tau}{2\sqrt{3\tau + C_1}} = -\frac{1}{6}\sqrt{3\tau + C_1} + C_4$$ $$z = \int \frac{\sqrt{p} - \sqrt{q}}{2} d\tau = \int \frac{\sqrt{3\tau + C_1} - \sqrt{C_2}}{2} d\tau = -\frac{1}{9}(-3\tau + C_1)^{3/2} - \frac{1}{2}\tau\sqrt{C_2} + C_5$$ $$\sqrt{3\tau + C_1} = 6(C_4 - x) \quad ; \quad \tau = 2\sqrt{C_2}(C_3 - y)$$ $$z = -24(C_4 - x)^3 - C_2(C_3 - y) + C_5$$

Answer: $z = 24(x - c_1)^3 + c_2y + c_3$; $c_2 = q$.

Reference:

E. A. Kuznetsov, D. A. Shapiro, Methods of Mathematical Physics, Part I, Chapter 2, 2011.

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