Question

The solution of the differential equation
\[ (D^3 - 5D^2 + 7D - 3)y = e^{-2x} \] is:

• A. $y = (c_1 + c_2 x)e^{-x} + c_3 e^{3x} - \dfrac{1}{45}e^{-2x}$
• B. $y = (c_1 + c_2 x)e^{x} + c_3 e^{-3x} + \dfrac{1}{45}e^{-2x}$
• C. $y = (c_1 + c_2 x)e^{-x} + c_3 e^{-3x} - \dfrac{1}{45}e^{-2x}$
• D. $y = (c_1 + c_2 x)e^{x} + c_3 e^{3x} - \dfrac{1}{45}e^{-2x}$

Hint:

Factor the auxiliary equation completely and check if the RHS is part of the CF. If not, directly assume $Ae^{ax}$ as PI.

Answer 1

The Correct Option is D

Step 1: Characteristic Equation of the Homogeneous Part 
The homogeneous equation is: \[ (D^3 - 5D^2 + 7D - 3)y = 0 \] Its characteristic equation is: \[ D^3 - 5D^2 + 7D - 3 = 0 \] Using rational root theorem, we try $D = 1$: \[ 1^3 - 5(1)^2 + 7(1) - 3 = 1 - 5 + 7 - 3 = 0 \] So $(D - 1)$ is a factor. Use polynomial division: \[ D^3 - 5D^2 + 7D - 3 = (D - 1)(D^2 - 4D + 3) \] Now factor the quadratic: \[ D^2 - 4D + 3 = (D - 1)(D - 3) \] So roots are: $D = 1$ (double root), $D = 3$\ Hence, complementary function: \[ y_c = (c_1 + c_2 x)e^{x} + c_3 e^{3x} \] Step 2: Particular Integral (PI) 
Given RHS is $e^{-2x}$, which is not part of the CF, so assume: \[ y_p = Ae^{-2x} \] Apply operator: \[ (D^3 - 5D^2 + 7D - 3)Ae^{-2x} = A(-2)^3 - 5A(-2)^2 + 7A(-2) - 3A = -8A - 20A - 14A - 3A = -45A \] So: \[ -45A = 1 \Rightarrow A = -\dfrac{1}{45} \] Particular Integral: $y_p = -\dfrac{1}{45}e^{-2x}$ 
Step 3: General Solution 
\[ y = y_c + y_p = (c_1 + c_2 x)e^{x} + c_3 e^{3x} - \dfrac{1}{45}e^{-2x} \]