Question

6. Find the integral curves of the following equations:

a) dx / x^2 - y^2 - yz = dy / x^2 - y^2 - zx = dz / z (x - y)

b) dx / x^2 + y^2 = dy / 2xy = dz / z (x + y)

c) Find the integral surface of the partial differential equation
(x - y) y^2 p + (y - x) x^2 q = (x^2 + y^2 ) z
through the curve xz = a^2, y = 0

Answer 1

Answer on Question #50900 – Math – Differential Calculus | Equations

a) Given:

$$\frac{dx}{x^2 - y^2 - yz} = \frac{dy}{x^2 - y^2} = \frac{dz}{z(x - y)}$$

Find: the integral curves of the equations

Solution

$$\frac{dx}{x^2 - y^2 - yz} = \frac{dy}{x^2 - y^2} \quad \Rightarrow \quad dx = dy - \frac{yz}{x^2 - y^2} \, dy$$ $$\frac{dy}{x^2 - y^2} = \frac{dz}{z(x - y)} \quad \Rightarrow \quad dy = \frac{x + y}{z} \, dz$$ $$\frac{dx}{x^2 - y^2 - yz} = \frac{dz}{z(x - y)} \quad \Rightarrow \quad dx = \frac{x + y}{z} \, dz - \frac{y}{x - y} \, dz$$

(2) and (3) $\Rightarrow$

$$dx = dy - \frac{y}{x - y} \, dz \quad \text{(4)} \Rightarrow$$

(1) and (4) $\Rightarrow$

$$\frac{y}{x - y} \, dz = \frac{yz}{x^2 - y^2} \, dy, \text{ divide both sides by } \frac{y}{x - y}$$ $$\Rightarrow \quad dz = \frac{z}{x + y} \, dy, \text{ separate variables}$$ $$\frac{dz}{z} = \frac{dy}{x + y}$$

Integrate both sides

$$Ln|z| = Ln|x + y| + Ln|c|$$

Apply properties of logarithmic function

$$|z| = |c(x + y)|,$$

leave signs of absolute values by means of choice of $c$,

$$z = c(x + y), \text{ where } c \text{ is arbitrary real constant}.$$

Answer: $z = c(x + y)$, $c = \text{const}$

b) Given:

$$\frac{dx}{x^2 + y^2} = \frac{dy}{2xy} = \frac{dz}{z(x + y)}$$

Find: the integral curves of the equations

Solution

$$\frac {d x}{x ^ {2} + y ^ {2}} = \frac {d y}{2 x y} \quad \Rightarrow \quad \frac {d y}{d x} = \frac {2 x y}{x ^ {2} + y ^ {2}}$$

it is a homogeneous differential equation so we replace $y = ux$

$$\frac {d y}{d x} = \left(u x\right) ^ {\prime} = u + u ^ {\prime} x = \frac {2 u x ^ {2}}{x ^ {2} + u ^ {2} x ^ {2}} = \frac {2 u}{1 + u ^ {2}}$$ $$u ^ {\prime} x = \frac {2 u}{1 + u ^ {2}} - u = \frac {2 u - u - u ^ {3}}{1 + u ^ {2}} = \frac {u - u ^ {3}}{1 + u ^ {2}}$$ $$\frac {d u}{d x} x = \frac {u - u ^ {3}}{1 + u ^ {2}} \quad \Rightarrow \quad \frac {1 + u ^ {2}}{u - u ^ {3}} d u = \frac {d x}{x}$$

and now we decompose on simple fractions

$$\frac {1 + u ^ {2}}{u (1 - u) (1 + u)} = \frac {A}{u} + \frac {B}{1 - u} + \frac {C}{1 + u}$$ $$\frac {A}{u} + \frac {B}{1 - u} + \frac {C}{1 + u} = \frac {A - A u ^ {2} + B u + B u ^ {2} + C u - C u ^ {2}}{u (1 - u) (1 + u)} = \frac {1 + u ^ {2}}{u (1 - u) (1 + u)} \quad \Rightarrow$$ $$\left\{ \begin{array}{l} - A + B - C = 1 \\ B + C = 0 \\ A = 1 \end{array} \right. \quad \Rightarrow \quad \left\{ \begin{array}{l} B - C = 2 \\ B = - C \\ A = 1 \end{array} \right. \quad \Rightarrow \quad \left\{ \begin{array}{l} 2 B = 2 \\ B = - C \\ A = 1 \end{array} \right. \quad \Rightarrow \quad \left\{ \begin{array}{l} A = 1 \\ B = 1 \\ C = - 1 \end{array} \right.$$ $$\Rightarrow \int \left(\frac {1}{u} + \frac {1}{1 - u} - \frac {1}{1 + u}\right) d u = \int \frac {d x}{x}$$ $$L n | u | - L n | 1 - u | - L n | 1 + u | = L n | x | + L n | c |$$ $$\frac {u}{1 - u ^ {2}} = x c$$

and now we return to $y$

$$\frac {\frac {y}{x}}{1 - \frac {y ^ {2}}{x ^ {2}}} = x c \Rightarrow$$ $$\frac {y}{x ^ {2} - y ^ {2}} = c \Rightarrow \quad x = \sqrt {\frac {y}{c} + y ^ {2}}$$ $$\frac {d y}{2 x y} = \frac {d z}{z (x + y)} \quad \Rightarrow \quad \frac {\sqrt {\frac {y}{c} + y ^ {2}} + y}{2 y \sqrt {\frac {y}{c} + y ^ {2}}} d y = \frac {d z}{z}$$ $$\int \left(\frac {d y}{2 y} + \frac {d y}{2 \sqrt {\frac {y}{c} + y ^ {2}}}\right) = \int \frac {d z}{z}$$

At first we calculate the integral

$$\int \frac {d y}{\sqrt {\frac {y}{c} + y ^ {2}}} = \int \frac {d y}{\sqrt {\left(y + \frac {1}{2 c}\right) ^ {2} - \frac {1}{4 c ^ {2}}}} = L n \left| y + \frac {1}{2 c} + \sqrt {\frac {y}{c} + y ^ {2}} \right|$$

So we obtain

$$\begin{array}{l} (1) \Rightarrow L n | y | + L n \left| y + \frac {1}{2 c} + \sqrt {\frac {y}{c} + y ^ {2}} \right| = L n | z | ^ {2} \\ z ^ {2} = y \left(y + \frac {1}{2 c} + \sqrt {\frac {y}{c} + y ^ {2}}\right) \\ \end{array}$$

Answer: $z^2 = y(x + y + \frac{1}{2c})$, $c = \text{const}$

c) Given:

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

where $p = \frac{\partial z}{\partial x}$ $q = \frac{\partial z}{\partial y}$

Find:

the integral surfaces through the curve $xz = a^2$, $y = 0$

Solution

$$\begin{array}{l} (x - y) y ^ {2} \frac {\partial z}{\partial x} + (y - x) x ^ {2} \frac {\partial z}{\partial y} = \left(x ^ {2} + y ^ {2}\right) z, z (x, 0) = \frac {a ^ {2}}{x} \\ \frac {d x}{(x - y) y ^ {2}} = \frac {d y}{(y - x) x ^ {2}} \text{ or } \frac {d z}{(x ^ {2} + y ^ {2}) z} = \frac {d y}{(y - x) x ^ {2}} \text{ or } \frac {d z}{(x ^ {2} + y ^ {2}) z} = \frac {d x}{(x - y) y ^ {2}} \\ \end{array}$$

In equation $\frac{dx}{(x - y)y^2} = \frac{dy}{(y - x)x^2}$ multiply both sides by $(x - y)$:

$$\frac {d x}{y ^ {2}} = - \frac {d y}{x ^ {2}}$$

So $x^{2}dx + y^{2}dy = 0$, which gives $x^{3} + y^{3} = C_{1}$.

We put $y = 0$ and $z = \frac{a^2}{x}$ to the partial differential equation, which yields

$$\begin{array}{l} - x ^ {3} \frac {\partial z}{\partial y} = x ^ {2} z \\ - x ^ {3} \frac {\partial z}{\partial y} = x a ^ {2} \\ x ^ {2} \frac {\partial z}{\partial y} = - a ^ {2} \\ \end{array}$$ $$\begin{array}{l} \frac{\partial z}{\partial y} = -\frac{a^{2}}{x^{2}} \\ z = -\frac{a^{2}}{x^{2}} y + \varphi(x) \\ xz = -\frac{a^{2}}{x} y + x \varphi(x) \end{array}$$

For $y = 0$ put $-\frac{a^2}{x} \cdot 0 + x\varphi(x) = a^2$, hence $\varphi(x) = \frac{a^2}{x}$

$$z = -\frac{a^{2}}{x^{2}} y + \frac{a^{2}}{x}$$

Answer: $z = -\frac{a^2}{x^2} y + \frac{a^2}{x}$

www.AssignmentExpert.com