We are tasked with finding a particular solution to the second-order ordinary differential equation (ODE):

$y'' + 9y = 6e^{3x} + 3\sin(3x)$

This is a non-homogeneous equation, and we will use the method of undetermined coefficients to find the particular solution.

Step 1: Solve the Homogeneous Equation

The corresponding homogeneous equation is:

$y'' + 9y = 0$

The characteristic equation is:

$r^2 + 9 = 0$

Solving for $r$:

$r = \pm 3i$

Thus, the general solution to the homogeneous equation is:

$y_h(x) = c_1 \cos(3x) + c_2 \sin(3x)$

Step 2: Solve for the Particular Solution

The non-homogeneous term is $6e^{3x} + 3\sin(3x)$. We need to find the particular solution $y_p(x)$ that fits this form.

• For the $6e^{3x}$ term, we assume a particular solution of the form: $y_{p1}(x) = Ae^{3x}$

• For the $3\sin(3x)$ term, since $\sin(3x)$ and $\cos(3x)$ already appear in the homogeneous solution, we need to multiply by $x$ to avoid overlap. Thus, we assume a particular solution of the form: $y_{p2}(x) = x(B \cos(3x) + C \sin(3x))$

So, the general form of the particular solution is:

$y_p(x) = Ae^{3x} + x(B \cos(3x) + C \sin(3x))$

Step 3: Determine the Coefficients

1. Substitute $y_{p1}(x) = Ae^{3x}$ into the ODE:

$y_{p1}'' + 9y_{p1} = 6e^{3x}$

First, compute the derivatives of $y_{p1}$: $y_{p1}' = 3Ae^{3x}, \quad y_{p1}'' = 9Ae^{3x}$

Substituting into the ODE:

$9Ae^{3x} + 9Ae^{3x} = 6e^{3x}$

$18A = 6 \quad \Rightarrow \quad A = \frac{1}{3}$

Thus, $y_{p1}(x) = \frac{1}{3}e^{3x}$.

2. Substitute $y_{p2}(x) = x(B \cos(3x) + C \sin(3x))$ into the ODE:

First, compute the derivatives of $y_{p2}$: $y_{p2}' = B \cos(3x) + C \sin(3x) + x(-3B \sin(3x) + 3C \cos(3x))$

$y_{p2}'' = -6B \sin(3x) + 6C \cos(3x) + x(-9B \cos(3x) - 9C \sin(3x))$

Substituting into the ODE:

$(-6B \sin(3x) + 6C \cos(3x) + x(-9B \cos(3x) - 9C \sin(3x))) + 9(x(B \cos(3x) + C \sin(3x))) = 3 \sin(3x)$

Simplifying:

$-6B \sin(3x) + 6C \cos(3x) + x(0) = 3 \sin(3x)$

Equating coefficients of $\sin(3x)$ and $\cos(3x)$:

• Coefficient of $\sin(3x)$: $-6B = 3 \quad \Rightarrow \quad B = -\frac{1}{2}$
• Coefficient of $\cos(3x)$: $6C = 0 \quad \Rightarrow \quad C = 0$

Thus, $y_{p2}(x) = -\frac{1}{2}x \cos(3x)$.

Step 4: General Solution

The general solution to the ODE is:

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

$y(x) = c_1 \cos(3x) + c_2 \sin(3x) + \frac{1}{3}e^{3x} - \frac{1}{2}x \cos(3x)$

This is the general solution to the given ODE.

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Would you like further details on any part?

Here are some related questions:

1. How does the method of undetermined coefficients compare with variation of parameters?
2. What is the role of the characteristic equation in solving homogeneous ODEs?
3. Why do we multiply by $x$ when terms appear in the homogeneous solution?
4. What are the limitations of using undetermined coefficients?
5. Can variation of parameters be applied to this same problem, and how?

Tip: Always check if any terms of the particular solution overlap with the homogeneous solution when using undetermined coefficients.