determine the general solution of the given differential equation: $y'' − 4y' + 13y = e^{2x}$ [closed]

$$ y(x) = y_h(x) + y_p(x), $$

where $y_h(x)$ is a solution of a homogeneous equation $y''-4y'+13y = 0$, and $y_p(x)$ is a particular solution.

To solve the equation $y''-4y'+13y = 0$, one have to solve a characteristic equation

$$ \lambda^2-4\lambda + 13 = 0, $$

which is a quadratic algebraic equation. The solution is $$ \lambda_{1, 2} = \frac{-(-4)\pm \sqrt{(-4)^2-4\cdot 1 \cdot 13}}{2\cdot 1} = 2\pm 3i \Rightarrow $$

$$ y_h(x) = C_1e^{2x}\cos(3x) + C_2e^{2x}\sin(3x). $$

The solution $y_p(x)$ can be found in a form $y_p(x) = Ae^{2x}$. Then,

$$ y_p'(x) = 2Ae^{2x}, \text{ and } y_p''(x) = 4Ae^{2x}. $$

Substituting $y_p(x), y_p'(x), \text{ and }y_p''(x)$ into the equation, one has

$$ 4Ae^{2x}-4\cdot 2Ae^{2x} + 13Ae^{2x} = e^{2x} \Leftrightarrow 9A = 1 \Leftrightarrow A = \frac{1}{9}, $$

and $y_p(x) = \frac{1}{9}e^{2x}$.

Finally, the general solution is $$ y(x) = y_h(x)+y_p(x) = C_1e^{2x}\cos(3x) + C_2e^{2x}\sin(3x) + \frac{1}{9}e^{2x}. $$