The given differential equation is:

$\frac{d^2 y}{dx^2} - 4\frac{dy}{dx} + 3y = \cos(x)$

This is a second-order linear differential equation with constant coefficients and a non-homogeneous term on the right-hand side ($\cos(x)$).

Steps to solve:

1. Solve the homogeneous equation:

   The corresponding homogeneous equation is: $\frac{d^2 y}{dx^2} - 4\frac{dy}{dx} + 3y = 0$ The characteristic equation is: $r^2 - 4r + 3 = 0$ Solving this quadratic equation: $r = \frac{4 \pm \sqrt{16 - 12}}{2} = \frac{4 \pm 2}{2}$ The roots are $r_1 = 3$ and $r_2 = 1$. Hence, the general solution to the homogeneous equation is: $y_h(x) = c_1 e^{3x} + c_2 e^{x}$

2. Find the particular solution:

   Since the non-homogeneous term is $\cos(x)$, we will try a particular solution of the form: $y_p(x) = A\cos(x) + B\sin(x)$ Differentiate $y_p(x)$: $\frac{dy_p}{dx} = -A\sin(x) + B\cos(x)$ $\frac{d^2 y_p}{dx^2} = -A\cos(x) - B\sin(x)$ Substituting $y_p(x)$, $\frac{dy_p}{dx}$, and $\frac{d^2 y_p}{dx^2}$ into the original equation: $(-A\cos(x) - B\sin(x)) - 4(-A\sin(x) + B\cos(x)) + 3(A\cos(x) + B\sin(x)) = \cos(x)$ Simplify: $(-A + 4B + 3A)\cos(x) + (-B - 4A + 3B)\sin(x) = \cos(x)$ This gives two equations by comparing coefficients: $(-A + 4B + 3A) = 1 \quad \text{(for the cosine terms)}$ $(-B - 4A + 3B) = 0 \quad \text{(for the sine terms)}$ Simplifying these: $2A + 4B = 1 \quad \text{(1)}$ $-4A + 2B = 0 \quad \text{(2)}$ From equation (2): $B = 2A$ Substituting $B = 2A$ into equation (1): $2A + 4(2A) = 1 \quad \Rightarrow \quad 10A = 1 \quad \Rightarrow \quad A = \frac{1}{10}$ Therefore, $B = 2A = \frac{2}{10} = \frac{1}{5}$.

   Thus, the particular solution is: $y_p(x) = \frac{1}{10} \cos(x) + \frac{1}{5} \sin(x)$

3. General solution:

   The general solution is the sum of the homogeneous and particular solutions: $y(x) = c_1 e^{3x} + c_2 e^{x} + \frac{1}{10} \cos(x) + \frac{1}{5} \sin(x)$

Would you like more details or have any questions?

Related questions:

1. How do you solve second-order differential equations with constant coefficients?
2. What is the method of undetermined coefficients in solving non-homogeneous differential equations?
3. How do you solve a quadratic characteristic equation in differential equations?
4. What happens when the characteristic equation has complex roots?
5. Can the method of variation of parameters be used to solve the same problem?

Tip:

For non-homogeneous differential equations, always look for the form of the particular solution based on the type of the non-homogeneous term (e.g., exponential, trigonometric, polynomial).