Question

Solve the following differential equations
i) (x^2) p + (y^2) q = (x + y ) z
ii) √p - √q + 3x = 0

Answer 1

Answer on Question #68974 – Math – Differential Equations

Question

Solve the following differential equation

i)

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

Solution

Let

$$p = \frac {\partial z}{\partial x}; \quad q = \frac {\partial z}{\partial y}.$$

We can present this quasilinear partial differential equation of the first order

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

as

$$V (x, y, z) = 0,$$

where

$$\frac {\partial z}{\partial x} = - \frac {\frac {\partial V}{\partial x}}{\frac {\partial V}{\partial z}},$$ $$\frac {\partial z}{\partial y} = - \frac {\frac {\partial V}{\partial y}}{\frac {\partial V}{\partial z}}.$$

Then

$$x ^ {2} \cdot \left(- \frac {\frac {\partial V}{\partial x}}{\frac {\partial V}{\partial z}}\right) + y ^ {2} \cdot \left(- \frac {\frac {\partial V}{\partial y}}{\frac {\partial V}{\partial z}}\right) = (x + y) z,$$ $$- x ^ {2} \frac {\partial V}{\partial x} - y ^ {2} \frac {\partial V}{\partial y} - (x + y) z \frac {\partial V}{\partial z} = 0,$$ $$x ^ {2} \frac {\partial V}{\partial x} + y ^ {2} \frac {\partial V}{\partial y} + (x + y) z \frac {\partial V}{\partial z} = 0.$$

The associated system of equations is

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

If $\frac{dx}{x^2} = \frac{dy}{y^2}$ , then $-\frac{1}{x} = -\frac{1}{y} + C_1$ and $u(x,y) = C_1 = \frac{1}{y} - \frac{1}{x}$ , hence $y = \frac{1}{C_1 + \frac{1}{x}}$ ,

$$y = \frac {x}{C _ {1} x + 1}$$

If

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

substituting for $y$ from (4) into (3) one gets

$$\frac {d x}{x ^ {2}} = \frac {d z}{(x + \frac {x}{C _ {1} x + 1}) z},$$ $$\frac {d x}{x} = \frac {d z}{\left(1 + \frac {1}{C _ {1} x + 1}\right) z},$$ $$\frac {d x}{x} = \frac {(C _ {1} x + 1)}{C _ {1} x + 2} \cdot \frac {d z}{z},$$ $$\frac {C _ {1} x + 2}{x \left(C _ {1} x + 1\right)} d x = \frac {d z}{z},$$ $$\int \frac {C _ {1} x + 2}{x \left(C _ {1} x + 1\right)} d x = \int \frac {d z}{z},$$ $$\int \left(\frac {2}{x} - \frac {C _ {1}}{C _ {1} x + 1}\right) d x = \ln | z | + \ln | C _ {2} |,$$ $$2 \ln | x | - \ln | C _ {1} x + 1 | = \ln | C _ {2} z |,$$ $$\frac {x ^ {2}}{C _ {1} x + 1} = C _ {2} z,$$ $$C _ {2} = \frac {x ^ {2}}{z (C _ {1} x + 1)},$$ $$C _ {2} = \frac {x ^ {2}}{z \left(\left(\frac {1}{y} - \frac {1}{x}\right) x + 1\right)}$$ $$C _ {2} = \frac {x ^ {2}}{z _ {y} ^ {2}},$$ $$v (x, y) = C _ {2} = \frac {x y}{z}.$$

The general integral is given by

$$F (u, v) = 0,$$ $$F \left(\frac {1}{y} - \frac {1}{x}, \frac {x y}{z}\right) = 0,$$ $$F \left(\frac {x - y}{x y}, \frac {x y}{z}\right) = 0,$$

where $F$ is an arbitrary function.

For example, if we take

$$F (u, v) = - C u v - 1 = - C \cdot \frac {x - y}{x y} \cdot \frac {x y}{z} - 1 = C \cdot \frac {y - x}{z} - 1 = 0,$$

then $z = C(y - x)$ is one of solutions of (1).

Function $z = 0$ is also a solution of equation (1) because $\frac{\partial z}{\partial x} = 0$, $\frac{\partial z}{\partial y} = 0$.

If $x = 0$, then $y^{2}\frac{\partial z}{\partial y} = yz$, $\frac{dz}{z} = \frac{dy}{y}$, $\frac{z}{y} = C_{1}$ and a solution of (1) is

$$(x, y, z) = (0, y, C _ {1} y) = s (0, 1, C _ {1}).$$

If $y = 0$, then $x^{2}\frac{\partial z}{\partial x} = xz$, $\frac{dz}{z} = \frac{dx}{x}$, $\frac{z}{x} = C_{2}$ and a solution of (1) is

$$(x, y, z) = (x, 0, C _ {2} x) = t (1, 0, C _ {2}).$$

If $y = -x$, then $z$ is arbitrary and a solution of (1) is $(x, y, z) = (x, -x, z) = (s, -s, t)$.

Answer: $F\left(\frac{x - y}{xy}, \frac{xy}{z}\right) = 0; z = C_1y, x = 0; z = C_2x, y = 0; y = -x.$

Question

Solve the following differential equation

ii)

$$\sqrt {p} - \sqrt {q} + 3 x = 0$$

Solution

$$F (x, y, z, p, q) = \sqrt {p} - \sqrt {q} + 3 x,$$ $$\frac {\partial F}{\partial x} = 3, \frac {\partial F}{\partial y} = 0, \frac {\partial F}{\partial z} = 0, \frac {\partial F}{\partial p} = \frac {1}{2 \sqrt {p}}, \frac {\partial F}{\partial q} = - \frac {1}{2 \sqrt {q}}.$$

The characteristic system of ordinary differential equations is

$$\frac {d x}{F _ {p}} = \frac {d y}{F _ {q}} = \frac {d z}{p F _ {p} + q F _ {q}} = - \frac {d p}{F _ {x} + p F _ {z}} = - \frac {d q}{F _ {y} + q F _ {z}}$$ $$\frac {d x}{\frac {1}{2 \sqrt {p}}} = \frac {d y}{- \frac {1}{2 \sqrt {q}}} = \frac {d z}{p \cdot \frac {1}{2 \sqrt {p}} + q \cdot \left(- \frac {1}{2 \sqrt {q}}\right)} = - \frac {d p}{3} = - \frac {d q}{0}$$ $$\frac {d x}{\frac {1}{2 \sqrt {p}}} = \frac {d y}{- \frac {1}{2 \sqrt {q}}} = \frac {d z}{\frac {\sqrt {p}}{2} - \frac {\sqrt {q}}{2}} = - \frac {d p}{3} = - \frac {d q}{0}$$

If

$$\frac {\frac {d y}{1}}{\frac {1}{2 \sqrt {q}}} = - \frac {d q}{0},$$ $$2 \sqrt {q} d y = \frac {d q}{0},$$ $$\frac {d q}{\sqrt {q}} = 0 d y,$$ $$2 \sqrt {q} = K,$$ $$\frac {\partial z}{\partial y} = q = C _ {1} ^ {2}$$

$(K, C_{1}$ are arbitrary constants),

hence

$$z = C _ {1} ^ {2} y + \varphi (x),$$ $$p = \frac {\partial z}{\partial x} = \varphi^ {\prime} (x)$$

Substituting (6), (7) into equation (5)

$$\sqrt {\varphi^ {\prime} (x)} - C _ {1} + 3 x = 0,$$ $$\sqrt {\varphi^ {\prime} (x)} = - 3 x + C _ {1}$$ $$\varphi^ {\prime} (x) = (- 3 x + C _ {1}) ^ {2} = 9 x ^ {2} - 6 x C _ {1} + C _ {1} ^ {2}$$ $$\varphi (x) = 9 \frac {x ^ {3}}{3} - 6 C _ {1} \frac {x ^ {2}}{2} + C _ {1} ^ {2} x + C _ {2} = 3 x ^ {3} - 3 C _ {1} x ^ {2} + C _ {1} ^ {2} x + C _ {2},$$

where $C_2$ is arbitrary constant.

Then

$$z = C _ {1} ^ {2} y + \varphi (x) = 3 x ^ {3} - 3 C _ {1} x ^ {2} + C _ {1} ^ {2} x + C _ {2} + C _ {1} ^ {2} y$$

If $\frac{\partial z}{\partial x} = p = 0$ , then it follows from (5) that

$\frac{\partial z}{\partial y} = q = 9x^2 \Rightarrow z = 9x^2y + \varphi(x) = 9C_1^2y + C_2$ (formula (8) already contains this solution).

If $\frac{\partial z}{\partial y} = q = 0$ , then it follows from (5) that

$\frac{\partial z}{\partial x} = p = 9x^2 \Rightarrow z = 3x^3 + \psi(y) = 3x^3 + c$ (formula (8) already contains this solution).

Answer: $z = 3x^{3} - 3C_{1}x^{2} + C_{1}^{2}x + C_{2} + C_{1}^{2}y$ .

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