Question

9. a) Find f ( y) so that equation f (y)dx − zx dy − xy ln y dz = 0 is integrable. Also obtain
the corresponding integral using Natani’s method.

b) Find the differential equation of the family of surfaces
phi [z (x + y)^2 , x^2 - y^2]=0

Answer 1

Answer on Question #51034 – Math – Differential Calculus

Question:

a) Find $f(y)$ so that equation $f(y)dx - zxdy - xy\ln ydz = 0$ is integrable. Also obtain the corresponding integral using Natani's method.

b) Find the differential equation of the family of surfaces $\Phi (z(x + y)^2,x^2 -y^2) = 0$

Solution:

a) Equation $f(y)dx - zxdy - xy\ln ydz = 0$ is integrable, if

$$X \cdot \operatorname {rot} X = 0$$ $$X = (f (y), - z x, - x y \ln y)$$

Let's calculate $\operatorname{rot} X$

$$\begin{array}{l} \operatorname {rot} X = \nabla \times X = \left| \begin{array}{c c c} \mathbf {i} & \mathbf {j} & \mathbf {k} \\ \frac {\partial}{\partial x} & \frac {\partial}{\partial y} & \frac {\partial}{\partial z} \\ f (y) & - z x & - x y \ln y \end{array} \right| \\ = \boldsymbol {i} \left(\frac {\partial}{\partial y} (- x y \ln y) - \frac {\partial}{\partial z} (- x z)\right) - \boldsymbol {j} \left(\frac {\partial}{\partial x} (- x y \ln y) - \frac {\partial}{\partial z} f (y)\right) \\ + \boldsymbol {k} \left(\frac {\partial}{\partial x} (- z x) - \frac {\partial}{\partial y} f (y)\right) = (- x \ln y) \boldsymbol {i} + (y \ln y) \boldsymbol {j} + (- z - f ^ {\prime} (y)) \boldsymbol {k} \\ \end{array}$$

Then

$$X \cdot \operatorname {rot} X = - x \ln y \cdot f (y) + x y \ln y f ^ {\prime} (y) = 0$$

Thus $x\ln y\cdot (yf'(y) - f(y)) = 0$, hence $yf^{\prime}(y) - f(y) = 0$

The solution of the last equation is determined by

$$\frac {d f}{f} = \frac {d y}{y} \rightarrow \ln | f (y) | = \ln | y | + \ln | A | \rightarrow f (y) = A \cdot y$$

where $A$ is a real constant. Finally, equation takes the form

$$A y d x - z x d y - x y \ln y d z = 0$$

Using Natani's method we firstly put $z = \text{const}$ and solve the obtained equation

$$A y d x = z x d y \rightarrow A \frac {d x}{x} = z \frac {d y}{y} \rightarrow A \ln x = z \ln y + F (z)$$

Now we let $x = 1$. Then equation takes the form

$$- z d y - y \ln y d z = 0$$

and

\begin{array}{l} F(z) = -z \cdot \ln y.

Thus,

$$z \, dy + y \ln y \, dz = 0 \quad \rightarrow \frac{dy}{y \ln y} + \frac{dz}{z} = 0 \quad \rightarrow \ln |\ln y| = -\ln |z| + \ln |C| \quad \rightarrow \\ \ln y = \frac{C}{z}$$

Therefore,

$$F(z) = -z \ln y = -z \cdot \frac{C}{z} = -C = K$$

(inother constant).

Finally,

$$A \ln x = z \ln y + K$$

where $K$ is an arbitrary real constant.

b) The general solution of first-order differential equation

$$\Phi(z(x + y)^2, x^2 - y^2) = 0$$

implies that

$$z(x + y)^2 = c_1 = \text{const}$$ $$x^2 - y^2 = c_2 = \text{const}$$

After differentiating these both equations we get

$$dz = -2z \cdot \frac{dx + dy}{x + y}$$ $$x \, dx = y \, dy$$

First-order differential equation

$$a \cdot \frac{\partial z}{\partial x} + b \cdot \frac{\partial z}{\partial y} = c$$

is equivalent to

$$\frac{dx}{a} = \frac{dy}{b} = \frac{dz}{c}$$

From $x \, dx = y \, dy$ and previous equation, we get

$$\frac{dx}{y} = \frac{dy}{x} \quad \rightarrow \quad a = y, \quad b = x$$

Putting $dx = \frac{y}{x} \, dy$ into $dz = -2z \cdot \frac{dx + dy}{x + y}$, we get

$$dz = -2z \cdot \frac{dy}{x}$$

Thus

$$\frac{dz}{c} = -\frac{2z}{c} \cdot \frac{dy}{x} = \frac{dy}{b} = \frac{dy}{x} \rightarrow c = -2z$$

Finally,

$$\frac{dx}{y} = \frac{dy}{x} = \frac{dz}{-2z}$$

or

$$y \cdot \frac{\partial z}{\partial x} + x \cdot \frac{\partial z}{\partial y} = -2z$$

Answer: $y \cdot \frac{\partial z}{\partial x} + x \cdot \frac{\partial z}{\partial y} = -2z$

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