Question

solve dy/dx = (y+x-2) /(y-x-4)

Answer 1

Answer on Question #44337 – Math - Differential Calculus | Equations

Solve $\mathrm{dy}/\mathrm{dx} = (y + x - 2)/(y - x - 4)$

Solution:

We make the change of variables:

$$\left\{ \begin{array}{l} x = \xi + x _ {0} \\ y = \eta + y _ {0} \end{array} \right.$$

where $x_0$ and $y_0$ are the solutions of linear system of equations

$$\left\{ \begin{array}{l} y _ {0} + x _ {0} - 2 = 0 \\ y _ {0} - x _ {0} - 4 = 0 \end{array} \right.$$

Rewrite the system in matrix form and solve it by Gaussian Elimination

$$\left[ \begin{array}{rrr} 1 & 1 & 2 \\ 1 & -1 & 4 \end{array} \right]$$

from 2 rows we subtract the 1-th row, multiplied respectively by 1

$$\left[ \begin{array}{rrr} 1 & 1 & 2 \\ 0 & -2 & 2 \end{array} \right]$$

divide the 2-th row by -2

$$\left[ \begin{array}{rrr} 1 & 1 & 2 \\ 0 & 1 & -1 \end{array} \right]$$

from 1 rows we subtract the 2-th row

$$\left[ \begin{array}{rrr} 1 & 0 & 3 \\ 0 & 1 & -1 \end{array} \right]$$

Answer:

$$\left\{ \begin{array}{l} x _ {0} = - 1 \\ y _ {0} = 3 \end{array} \right.$$

For new variables we have $dx = d\xi$, $dy = d\eta$ and $\frac{d\eta}{d\xi} = \frac{(\eta + \xi)}{(\eta - \xi)}$.

We make the change of variable $z = \eta / \xi$, then $\eta = \xi * z$ and $\frac{d\eta}{d\xi} = z + \xi * \frac{dz}{d\xi}$.

Now for variable $z$ we obtain differential equation:

$$z + \xi * dz / d\xi = (z + 1) / (z - 1)$$

collecting terms containing $z$ we obtained equation with separable variables $\xi * dz / d\xi = (-z^2 + 2z + 1) / (z - 1)$.

Separating variables

$$\frac {z - 1}{- z ^ {2} + 2 z + 1} * d z = \frac {d \xi}{\xi}.$$

Integrating the left-hand side of the equality with respect to $z$ and right side with respect to $\xi$:

$$\int \frac {z - 1}{- z ^ {2} + 2 z + 1} d z = \int \frac {- 2 z + 2}{- 2 (- z ^ {2} + 2 z + 1)} d z = \int \frac {d (- z ^ {2} + 2 z + 1)}{- 2 (- z ^ {2} + 2 z + 1)} = - \frac {1}{2} * \ln | - z ^ {2} + 2 z + 1 | + C,$$ $$\int \frac {d \xi}{\xi} = \ln | \xi | + C$$

We have

$$- \frac {1}{2} * \ln | - z ^ {2} + 2 z + 1 | = \ln | \xi | + \ln C.$$

(we denote arbitrary constant C by lnC)

Using the properties of logarithms $\ln (x)$:

$$\ln \left| \frac {1}{\sqrt {- z ^ {2} + 2 z + 1}} \right| = \ln | \xi * C |$$

from here obtained

$$\frac {1}{\sqrt {- z ^ {2} + 2 z + 1}} = \xi * C$$

or rewrite equality by $C_1 = (-z^2 + 2z + 1) * \xi^2$, where $C_1$ arbitrary constant.

Back to old variables

$$Z = \frac {\eta}{\xi} = \frac {y - y _ {0}}{x - x _ {0}} = \frac {y - 3}{x + 1}$$

We have

$$\begin{array}{l} C _ {1} = (x + 1) ^ {2} * \left(- \left(\frac {y - 3}{x + 1}\right) ^ {2} + 2 \left(\frac {y - 3}{x + 1}\right) + 1\right) = - (y - 3) ^ {2} + 2 (y - 3) (x + 1) + \\ + (x + 1) ^ {2} = - y ^ {2} + 2 x y + x ^ {2} + 8 y - 4 x - 14. \\ \end{array}$$

denote arbitrary constant $C_1 + 14$ by C, then

**Answer:**

$-y^{2} + 2xy + x^{2} + 8y - 4x = C$, where C arbitrary constant, is a solution of differential equation $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{(y + x - 2)}{(y - x - 4)}$.

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