Let's solve the problem step by step: ### Given: The characteristic equation for a differential equation is \[ (r-1)^2(r-2)^2 = 0 \] #### **Step 1: Find the roots of the characteristic equation** \[ (r-1)^2(r-2)^2 = 0 \] - Roots: \( r = 1 \) (multiplicity 2), \( r = 2 \) (multiplicity 2) #### **Step 2: Write the general solution** For repeated roots, the general solution is: \[ y(x) = (C_1 + C_2 x)e^{r_1 x} + (C_3 + C_4 x)e^{r_2 x} \] So, \[ y(x) = (C_1 + C_2 x)e^{x} + (C_3 + C_4 x)e^{2x} \] #### **Step 3: Write the differential equation corresponding to the characteristic equation** Given characteristic equation: \[ (r-1)^2(r-2)^2 = 0 \] The corresponding differential operator is: \[ \left(D-1\right)^2\left(D-2\right)^2 y = 0 \] Expanding: First, \((D-1)^2 = D^2 - 2D + 1\) Second, \((D-2)^2 = D^2 - 4D + 4\) So, \[ \left(D^2 - 2D + 1\right)\left(D^2 - 4D + 4\right)y = 0 \] Now, expand \((D^2 - 2D + 1)(D^2 - 4D + 4)\): Multiply out: \[ = D^2(D^2 - 4D + 4) - 2D(D^2 - 4D + 4) + 1(D^2 - 4D + 4) \] \[ = (D^4 - 4D^3 + 4D^2) - (2D^3 - 8D^2 + 8D) + (D^2 - 4D + 4) \] \[ = D^4 - 4D^3 + 4D^2 - 2D^3 + 8D^2 - 8D + D^2 - 4D + 4 \] \[ = D^4 - 6D^3 + 13D^2 - 12D + 4 \] So, the differential equation is: \[ y^{(4)} - 6y''' + 13y'' - 12y' + 4y = 0 \] #### **Step 4: Match with the given options** Look at the options in the image and compare with the obtained equation. None of the boxed options directly match, but our derived equation is correct. --- ### **Final Answers** #### a. The differential equation is: \[ \boxed{y^{(4)} - 6y''' + 13y'' - 12y' + 4y = 0} \] #### b. The general solution is: \[ \boxed{y(x) = (C_1 + C_2x) e^x + (C_3 + C_4x) e^{2x}} \] Let me know if you have any questions!

Let's carefully analyze each part of the problem, providing detailed calculations, explanations, and comparing each option to determine the correct differential equation and its solution. --- ### Given Information: - The **characteristic equation** is: \[ (r - 1)^2(r - 2)^2 = 0 \] which implies roots: - \( r = 1 \) with multiplicity 2 - \( r = 2 \) with multiplicity 2 --- ## Part 1: Find the **differential equation** corresponding to this characteristic equation ### Step 1: Roots and multiplicities - Roots: \( r = 1 \) (multiplicity 2), \( r = 2 \) (multiplicity 2) ### Step 2: Write the **general solution** form For repeated roots, the general solution is: \[ y(x) = \left( C_1 + C_2 x \right) e^{r_1 x} + \left( C_3 + C_4 x \right) e^{r_2 x} \] which becomes: \[ \boxed{ y(x) = \left( C_1 + C_2 x \right) e^{x} + \left( C_3 + C_4 x \right) e^{2x} } \] ### Step 3: Construct the **differential operator** The characteristic equation factors as: \[ (r - 1)^2 (r - 2)^2 = 0 \] which corresponds to the differential operator: \[ \left( D - 1 \right)^2 \left( D - 2 \right)^2 y = 0 \] Let's expand this operator step-by-step. --- ## Part 2: Expand the differential operator ### Step 4: Expand \((D - 1)^2\) \[ (D - 1)^2 = D^2 - 2D + 1 \] ### Step 5: Expand \((D - 2)^2\) \[ (D - 2)^2 = D^2 - 4D + 4 \] ### Step 6: Multiply these two operators \[ (D^2 - 2D + 1)(D^2 - 4D + 4) \] Distribute term-by-term: \[ = D^2(D^2 - 4D + 4) - 2D(D^2 - 4D + 4) + 1(D^2 - 4D + 4) \] Calculate each: 1. \( D^2 \cdot D^2 = D^4 \) 2. \( D^2 \cdot (-4D) = -4 D^3 \) 3. \( D^2 \cdot 4 = 4 D^2 \) 4. \( -2D \cdot D^2 = -2 D^3 \) 5. \( -2D \cdot (-4 D) = +8 D^2 \) 6. \( -2D \cdot 4 = -8 D \) 7. \( 1 \cdot D^2 = D^2 \) 8. \( 1 \cdot (-4 D) = -4 D \) 9. \( 1 \cdot 4 = 4 \) Now, sum all these: \[ D^4 - 4 D^3 + 4 D^2 - 2 D^3 + 8 D^2 - 8 D + D^2 - 4 D + 4 \] Combine like terms: - \( D^4 \) - \( -4 D^3 - 2 D^3 = -6 D^3 \) - \( 4 D^2 + 8 D^2 + D^2 = 13 D^2 \) - \( -8 D - 4 D = -12 D \) - Constant: \( 4 \) **Resulting differential operator:** \[ \boxed{ D^4 - 6 D^3 + 13 D^2 - 12 D + 4 } \] --- ## **Final differential equation:** \[ \boxed{ \boxed{ y^{(4)} - 6 y''' + 13 y'' - 12 y' + 4 y = 0 } } \] --- ## Part 3: Find the **general solution** The roots: - \( r = 1 \), multiplicity 2 - \( r = 2 \), multiplicity 2 Corresponding solution components: \[ y(x) = (A + B x) e^{x} + (C + D x) e^{2x} \] where \( A, B, C, D \) are arbitrary constants. --- ## **Summary:** | Part | Details | Final Answer | |---------|--------------------------------------------------------------|--------------------------------------------------------------| | **a. Differential Equation** | Derived from roots \( r=1,2 \) with multiplicities 2 | \(\boxed{ y^{(4)} - 6 y''' + 13 y'' - 12 y' + 4 y = 0 } \) | | **b. General Solution** | Based on roots and multiplicities | \(\boxed{ y(x) = (C_1 + C_2 x) e^{x} + (C_3 + C_4 x) e^{2x} } \) | --- ## **Matching with provided options:** **Check options in the image:** - The **differential equation** corresponds to the expanded form: \( y^{(4)} - 6 y''' + 13 y'' - 12 y + 4 y = 0 \) - The **solution** matches the form: \((C_1 + C_2 x) e^{x} + (C_3 + C_4 x) e^{2x}\) --- ### **Note:** - If there are options in the question, pick the one matching the differential equation and solution form derived here. - The key steps involve expanding the characteristic roots into the differential operator and forming the general solution based on multiplicities. --- **Let me know if you'd like me to analyze specific options or clarify any step further!**