We are tasked to solve the non-homogeneous differential equation
$y'' - 2y' + y = 4xe^{3x}$ using the method of undetermined coefficients with initial conditions $y(0) = 2$ and $y'(0) = 1$.

Step 1: Solve the homogeneous part

The homogeneous part of the equation is: $y'' - 2y' + y = 0.$

The corresponding characteristic equation is: $r^2 - 2r + 1 = 0.$

Solving for $r$, we factorize: $(r - 1)^2 = 0 \implies r = 1 \ \text{(double root)}.$

Thus, the general solution of the homogeneous equation is: $y_h(x) = C_1e^{x} + C_2xe^{x},$ where $C_1$ and $C_2$ are constants.

Step 2: Solve the particular solution $y_p$

The non-homogeneous term is $4xe^{3x}$. Since $e^{3x}$ appears and is not a solution of the homogeneous equation, we choose a particular solution of the form: $y_p(x) = (Ax^2 + Bx)e^{3x}.$

Finding $y_p'(x)$ and $y_p''(x)$:

1. First derivative $y_p'(x)$: $y_p'(x) = \left(2Ax + B + 3Ax^2 + 3Bx\right)e^{3x} = (3Ax^2 + (2A + 3B)x + B)e^{3x}.$

2. Second derivative $y_p''(x)$: $y_p''(x) = \left[ 6Ax + (2A + 3B) + 3(3Ax^2 + (2A + 3B)x + B) \right]e^{3x}.$ Simplify term by term: $y_p''(x) = \left( 9Ax^2 + (6A + 9B)x + (2A + 3B) \right)e^{3x}.$

Substituting into the original equation:

Substitute $y_p$, $y_p'$, and $y_p''$ into the original equation: $y_p'' - 2y_p' + y_p = 4xe^{3x}.$

Substitute each expression: $\left( 9Ax^2 + (6A + 9B)x + (2A + 3B) \right)e^{3x} - 2\left( 3Ax^2 + (2A + 3B)x + B \right)e^{3x} + \left( Ax^2 + Bx \right)e^{3x}.$

Combine like terms:

• $x^2$-terms: $9A - 6A + A = 4A$,
• $x$-terms: $(6A + 9B) - 2(2A + 3B) + B = 6A + 9B - 4A - 6B + B = 2A + 4B$,
• Constant terms: $(2A + 3B) - 2B = 2A - B$.

Equating coefficients of $x^2$, $x$, and the constant on the left to those on the right-hand side ($4xe^{3x}$):

• $x^2$-term: $4A = 0 \implies A = 0$,
• $x$-term: $2A + 4B = 4 \implies 4B = 4 \implies B = 1$,
• Constant term: $2A - B = 0 \implies -B = 0 \quad \text{(consistent)}$.

Thus, the particular solution is: $y_p(x) = Bx e^{3x} = x e^{3x}.$

Step 3: General solution

The general solution $y(x)$ is the sum of the homogeneous and particular solutions: $y(x) = y_h(x) + y_p(x).$ Substitute: $y(x) = C_1e^{x} + C_2xe^{x} + xe^{3x}.$

Step 4: Apply initial conditions

Condition 1: $y(0) = 2$

Substitute $x = 0$ into $y(x)$: $y(0) = C_1e^{0} + C_2(0)e^{0} + (0)e^{3\cdot 0} = C_1.$ Thus: $C_1 = 2.$

Condition 2: $y'(0) = 1$

First, compute $y'(x)$: $y'(x) = C_1e^{x} + C_2e^{x} + C_2xe^{x} + e^{3x} + 3xe^{3x}.$ Simplify: $y'(x) = C_1e^{x} + C_2e^{x} + C_2xe^{x} + e^{3x} + 3xe^{3x}.$ At $x = 0$: $y'(0) = C_1e^{0} + C_2e^{0} + (0) + e^{3\cdot 0} + 3(0) = C_1 + C_2 + 1.$ Substitute $C_1 = 2$ and $y'(0) = 1$: $1 = 2 + C_2 + 1 \implies C_2 = -2.$

Final Solution:

The solution to the given differential equation is: $y(x) = 2e^{x} - 2xe^{x} + xe^{3x}.$

Conclusion

Do you need further clarification on any steps? Let me know!

Related Questions:

1. How does the method of undetermined coefficients differ from variation of parameters?
2. How do repeated roots in the characteristic equation affect the general solution?
3. Can you solve this equation using the Laplace Transform method?
4. How do you check if a solution satisfies the initial value problem?
5. What happens if the non-homogeneous term overlaps with the homogeneous solution?

Tip:

Always simplify derivatives carefully when solving non-homogeneous equations to avoid algebraic errors.