To solve the differential equation $(D^2 + 4)y = x^2 \cos(2x)$, where $D$ denotes differentiation with respect to $x$, let's proceed with the method of undetermined coefficients.

Step 1: Solve the Homogeneous Equation

The associated homogeneous equation is: $(D^2 + 4)y = 0$ This can be rewritten as: $y'' + 4y = 0$ The characteristic equation is: $r^2 + 4 = 0$ Solving for $r$, we get: $r = \pm 2i$ Thus, the complementary solution $y_c$ is: $y_c = C_1 \cos(2x) + C_2 \sin(2x)$

Step 2: Find the Particular Solution

Since the right-hand side of the differential equation is $x^2 \cos(2x)$, we can try a particular solution of the form: $y_p = x^2(A \cos(2x) + B \sin(2x))$ This form is chosen because $x^2 \cos(2x)$ and $x^2 \sin(2x)$ match the type of function on the right-hand side, and we multiply by $x^2$ because $\cos(2x)$ and $\sin(2x)$ are solutions of the homogeneous equation.

1. Differentiate $y_p$: Calculate $y_p'$ and $y_p''$, and substitute into the left-hand side of the original equation.

2. Equate terms with powers of $x$ on both sides of the equation to solve for the coefficients $A$ and $B$.

After calculating, you'll get values for $A$ and $B$ that make the equation true.

Step 3: General Solution

The general solution to the differential equation is: $y = y_c + y_p = C_1 \cos(2x) + C_2 \sin(2x) + x^2(A \cos(2x) + B \sin(2x))$ This is the complete solution.

Would you like the detailed steps for finding $A$ and $B$?

Related Questions

1. How do we apply the method of undetermined coefficients to other forms of non-homogeneous terms?
2. Can this method be used when the non-homogeneous term involves exponential functions?
3. What if the characteristic equation has real roots rather than complex roots?
4. How does the method of variation of parameters differ from undetermined coefficients?
5. In what scenarios do we need to use higher powers of $x$ in our particular solution guess?

Tip

Always check if the form of your particular solution overlaps with the complementary solution to decide on multiplying by a higher power of $x$.