Answer to Question #105436 in Differential Equations for Zoya

Question

Solve differential equations dy/dx=1+(dy/dx)^2.

Answer 1

Task: $\frac{dy}{dx} = 1 + (\frac{dy}{dx})^2$ .

Solution: It is convinient to use new variable: $f = \frac{dy}{dx}$ (1).

By substitution (1) it is possible to transform initial equation in the following form:

$f^2 -f + 1 = 0$

This is an ordinary quadratic equations with the following solution:

$D = (-1)^2 - 4 \cdot 1 \cdot 1 = -3 \\ f_{1,2} = \frac{1 \pm \sqrt{D}}{2} = \frac{1 \pm \sqrt{3}i}{2} \\ f_1 = \frac{1}{2} + \frac{\sqrt{3}}{2}i,\ f_2 = \frac{1}{2} - \frac{\sqrt{3}}{2}i$

where $i^2 = -1$ — imaginary unit.

By using reversed transform of (1) we can get the solutions:

$y_{1,2} = \int f_{1, 2} dx$

To integrate complex functions it is possible to use the feature of continous contour integrals:

$\int (u(x) + iv(x))dx = \int u(x) dx + i\int v(x)dx$

Thus, the solutions are:

$y_1(x) = \int f_1 dx = \int (\frac{1}{2} + \frac{\sqrt{3}}{2}i)dx = \int \frac{1}{2}dx + i\int \frac{\sqrt{3}}{2}dx=(\frac{1}{2} + \frac{\sqrt{3}}{2}i)x + C_1 \\ y_2(x) = \int f_2 dx = \int (\frac{1}{2} - \frac{\sqrt{3}}{2}i)dx = \int \frac{1}{2}dx - i\int \frac{\sqrt{3}}{2}dx=(\frac{1}{2} - \frac{\sqrt{3}}{2}i)x + C_2$

So, the final solution is:

$y(x) = (\frac{1}{2} \pm \frac{\sqrt{3}}{2}i)x + C$

Answer: $y(x) = (\frac{1}{2} \pm \frac{\sqrt{3}}{2}i)x + C$