Find the general solution of the homogeneous differential equations (D² - 3D - 10) y=0, y(0)=4, y'(0)=2

The given differential equation is $(D^2 - 3D - 10)y = 0$.
The characteristic equation is obtained by replacing $D$ with $r$:

$$r^2 - 3r - 10 = 0$$

Factor the quadratic equation:

$$(r-5)(r+2) = 0$$

The roots are $r_1 = 5$ and $r_2 = -2$. These are distinct real roots.

For distinct real roots $r_1$ and $r_2$, the general solution is of the form $y(x) = c_1 e^{r_1 x} + c_2 e^{r_2 x}$.
Substituting the roots:

$$y(x) = c_1 e^{5x} + c_2 e^{-2x}$$

Substitute $x=0$ and $y=4$ into the general solution:

$$4 = c_1 e^{5(0)} + c_2 e^{-2(0)}$$ $$4 = c_1 e^0 + c_2 e^0$$ $$4 = c_1 (1) + c_2 (1)$$ $$c_1 + c_2 = 4 \quad (*)$$

Differentiate $y(x)$ with respect to $x$:

$$y'(x) = \frac{d}{dx}(c_1 e^{5x} + c_2 e^{-2x})$$ $$y'(x) = 5c_1 e^{5x} - 2c_2 e^{-2x}$$

Substitute $x=0$ and $y'=2$ into the derivative:

$$2 = 5c_1 e^{5(0)} - 2c_2 e^{-2(0)}$$ $$2 = 5c_1 e^0 - 2c_2 e^0$$ $$2 = 5c_1 (1) - 2c_2 (1)$$ $$5c_1 - 2c_2 = 2 \quad (**)$$

We have the system:

1. $c_1 + c_2 = 4$
2. $5c_1 - 2c_2 = 2$

From equation (1), $c_2 = 4 - c_1$. Substitute this into equation (2):

$$5c_1 - 2(4 - c_1) = 2$$ $$5c_1 - 8 + 2c_1 = 2$$ $$7c_1 = 10$$ $$c_1 = \frac{10}{7}$$

Now, substitute $c_1$ back into $c_2 = 4 - c_1$:

$$c_2 = 4 - \frac{10}{7}$$ $$c_2 = \frac{28}{7} - \frac{10}{7}$$ $$c_2 = \frac{18}{7}$$

The particular solution is:

$$y(x) = \frac{10}{7} e^{5x} + \frac{18}{7} e^{-2x}$$