Question

The Particular integral of \((D^3 - 6D^2 + 11D - 6)y = e^{-2x}\) is:

• A. \( \frac{e^{2x}}{30} \)
• B. \( \frac{e^{-2x}}{60} \)
• C. \( -\frac{e^{-2x}}{60} \)
• D. \( \frac{e^{3x}}{50} \)

Hint:

Always check if the exponent in the non-homogeneous term is a root of the auxiliary equation. If it is, the form of the particular integral needs to be modified by multiplying by powers of \( x \). In this case, \( -2 \) is not a root (\( 1, 2, 3 \)), so the simple form \( A e^{-2x} \) works.

Answer 1

The Correct Option is C

Step 1: Identify the differential equation and the form of the non-homogeneous term.
The given linear non-homogeneous differential equation with constant coefficients is: \[ (D^3 - 6D^2 + 11D - 6)y = e^{-2x} \] Here, \( D = \frac{d}{dx} \). The non-homogeneous term is of the form \( f(x) = e^{ax} \), where \( a = -2 \). Step 2: Find the auxiliary equation and its roots (for the complementary function, although not strictly needed for the particular integral in this case).
The auxiliary equation is: \[ m^3 - 6m^2 + 11m - 6 = 0 \] By inspection, \( m = 1 \) is a root: \( 1 - 6 + 11 - 6 = 0 \).
Dividing by \( (m - 1) \): \( (m^2 - 5m + 6) = 0 \).
Factoring the quadratic: \( (m - 2)(m - 3) = 0 \).
The roots are \( m_1 = 1, m_2 = 2, m_3 = 3 \). These are distinct, so the complementary function is \( y_c = c_1 e^x + c_2 e^{2x} + c_3 e^{3x} \).
Step 3: Determine the form of the particular integral.
Since the non-homogeneous term is \( e^{-2x} \) and \( a = -2 \) is not a root of the auxiliary equation, the particular integral \( y_p \) will be of the form: \[ y_p = A e^{-2x} \] where \( A \) is a constant to be determined. Step 4: Substitute \( y_p \) and its derivatives into the original differential equation.
First, find the derivatives of \( y_p \): \[ y_p' = -2A e^{-2x} = Dy_p \] \[ y_p'' = 4A e^{-2x} = D^2 y_p \] \[ y_p''' = -8A e^{-2x} = D^3 y_p \] Substitute these into the differential equation: \[ (-8A e^{-2x}) - 6(4A e^{-2x}) + 11(-2A e^{-2x}) - 6(A e^{-2x}) = e^{-2x} \] Step 5: Solve for the constant \( A \).
Divide both sides by \( e^{-2x} \): \[ -8A - 24A - 22A - 6A = 1 \] \[ (-8 - 24 - 22 - 6)A = 1 \] \[ -60A = 1 \] \[ A = -\frac{1}{60} \] Step 6: Write the particular integral \( y_p \).
Substitute the value of \( A \) back into the form of \( y_p \): \[ y_p = -\frac{1}{60} e^{-2x} = -\frac{e^{-2x}}{60} \] Step 7: Select the correct answer.
The particular integral is \( -\frac{e^{-2x}}{60} \), which corresponds to option 3.