To solve the given differential equation using the method of variation of parameters, let us proceed step-by-step:

The Differential Equation

$y'' + 6y' + 9y = x^{-3}e^{-3x}, \quad x > 0.$

Step 1: Solve the Homogeneous Equation

The homogeneous equation is: $y'' + 6y' + 9y = 0.$

This is a second-order linear differential equation. The characteristic equation is: $r^2 + 6r + 9 = 0.$

Factoring: $(r + 3)^2 = 0.$

Thus, the roots are: $r = -3 \quad \text{(repeated root)}.$

The general solution to the homogeneous equation is: $y_h = c_1e^{-3x} + c_2xe^{-3x}.$

Step 2: Apply Variation of Parameters

The particular solution $y_p$ is found using the variation of parameters method. Assume: $y_p = u_1(x)e^{-3x} + u_2(x)xe^{-3x}.$

Here, $u_1(x)$ and $u_2(x)$ are functions to be determined.

Step 2.1: Derivatives of $y_p$

$y_p' = u_1'(x)e^{-3x} + u_1(x)(-3e^{-3x}) + u_2'(x)xe^{-3x} + u_2(x)(e^{-3x} - 3xe^{-3x}),$ $y_p' = (u_1'e^{-3x} + u_2'xe^{-3x}) - 3(u_1e^{-3x} + u_2xe^{-3x}),$ $y_p' = u_1'e^{-3x} + u_2'xe^{-3x} - 3y_p.$

The second derivative is: $y_p'' = u_1''e^{-3x} + u_1'(-3e^{-3x}) + u_2''xe^{-3x} + u_2'(e^{-3x} - 3xe^{-3x}) - 3(u_1'e^{-3x} + u_2'xe^{-3x} - 3y_p).$

Step 2.2: Simplify Equations

To simplify computation, we enforce the following condition: $u_1'e^{-3x} + u_2'xe^{-3x} = 0.$

This simplifies the first derivative to: $y_p' = -3(u_1e^{-3x} + u_2xe^{-3x}).$

Substitute into the original equation: $y_p'' + 6y_p' + 9y_p = x^{-3}e^{-3x}.$

After simplifications, the equations for $u_1'$ and $u_2'$ become: $u_1'(x)e^{-3x} + u_2'(x)xe^{-3x} = 0, \quad \text{(1)}$ $-3u_1'(x)e^{-3x} - 3u_2'(x)xe^{-3x} = x^{-3}e^{-3x}. \quad \text{(2)}$

Step 2.3: Solve for $u_1'$ and $u_2'$

From equation (1): $u_1'(x) + u_2'(x)x = 0 \implies u_1'(x) = -u_2'(x)x.$

Substitute $u_1'(x) = -u_2'(x)x$ into equation (2): $-3(-u_2'(x)x)e^{-3x} - 3u_2'(x)xe^{-3x} = x^{-3}e^{-3x}.$ $3u_2'(x)xe^{-3x} - 3u_2'(x)xe^{-3x} = x^{-3}e^{-3x}.$ $u_2'(x) = x^{-3}.$

Now substitute $u_2'(x)$ into $u_1'(x) = -u_2'(x)x$: $u_1'(x) = -x^{-3}x = -x^{-2}.$

Step 2.4: Integrate for $u_1(x)$ and $u_2(x)$

$u_2(x) = \int x^{-3} dx = -\frac{1}{2}x^{-2}.$ $u_1(x) = \int -x^{-2} dx = x^{-1}.$

Step 3: Write the Particular Solution

Substitute $u_1(x)$ and $u_2(x)$ into $y_p$: $y_p = u_1(x)e^{-3x} + u_2(x)xe^{-3x},$ $y_p = x^{-1}e^{-3x} - \frac{1}{2}x^{-2}xe^{-3x}.$ $y_p = x^{-1}e^{-3x} - \frac{1}{2}x^{-1}e^{-3x}.$ $y_p = \frac{1}{2}x^{-1}e^{-3x}.$

Step 4: General Solution

The general solution is: $y = y_h + y_p,$ $y = c_1e^{-3x} + c_2xe^{-3x} + \frac{1}{2}x^{-1}e^{-3x}.$

This is the required solution.

Would you like further explanation on any step or related concepts?

Related Questions:

1. How does the method of variation of parameters differ from undetermined coefficients?
2. Can you explain why we use $u_1'e^{-3x} + u_2'xe^{-3x} = 0$ as a simplifying assumption?
3. How would the solution change if the forcing term was $e^{-3x}$ instead of $x^{-3}e^{-3x}$?
4. Why do repeated roots in the characteristic equation result in solutions involving $x$?
5. What are the limitations of variation of parameters in solving differential equations?

Tip:

Always simplify intermediate derivatives when using variation of parameters to reduce algebraic complexity!