Question

Find domain and range of (answers should be subsets of R):

f(x)= 3/2x -1

Answer 1

Answer on Question #76040 – Math – Differential Equations

Question

Find the general integral of the equation

$$(x - y)p + (y - x - z)q = z$$

and particular solution through the circle

$$z = 1, x^2 + y^2 = 1$$

Solution

Auxiliary equations:

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

Then

$$\frac{dx + dy + dz}{x - y + y - x - z + z} = \frac{d(x + y + z)}{0}$$ $$d(x + y + z) = 0$$

The $1^{\mathrm{st}}$ solution:

$$\varphi_1 = x + y + z = c_1$$ $$\frac{dx - dy + dz}{x - y - y + x + z + z} = \frac{dz}{z}$$ $$\frac{d(x - y + z)}{x - y + z} = \frac{2dz}{z}$$ $$d\left(\ln \frac{x - y + z}{z^2}\right) = 0$$

The $2^{\mathrm{nd}}$ solution:

$$\varphi_2 = \frac{x - y + z}{z^2} = c_2$$

The general solution:

$$F(\varphi_1, \varphi_2) = F\left(x + y + z, \frac{x - y + z}{z^2}\right)$$

For $z = 1$:

$$\varphi_1 = x + y + z = x + y + 1 \Rightarrow x = \frac{\varphi_1 + \varphi_2}{2} - 1$$ $$\varphi_2 = \frac{x - y + z}{z^2} = x - y + 1 \Rightarrow y = \frac{\varphi_1 - \varphi_2}{2}$$ $$\left\{ \begin{array}{c} x = \frac{\varphi_1 + \varphi_2}{2} - 1 \\ y = \frac{\varphi_1 - \varphi_2}{2} \end{array} \right.$$

For $x^2 + y^2 = 1$:

$$\left(\frac{\varphi_1 + \varphi_2}{2} - 1\right)^2 + \left(\frac{\varphi_1 - \varphi_2}{2}\right)^2 = 1$$ $$\left(\frac{\varphi_1 + \varphi_2}{2}\right)^2 - (\varphi_1 + \varphi_2) + 1 + \left(\frac{\varphi_1 - \varphi_2}{2}\right)^2 = 1$$ $$\frac{\varphi_1^2 + \varphi_2^2}{2} - (\varphi_1 + \varphi_2) = 0$$ $$\varphi_1(\varphi_1 - 2) + \varphi_2(\varphi_2 - 2) = 0$$

Answer:

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

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