Let's break down the problem step by step: ### 1. **Given Equation:** \[ y^{(3)} - 9y' = te^{-3t} + 4\cos 3t \] ### 2. **Auxiliary Equation for the Homogeneous Part:** The homogeneous part is: \[ y^{(3)} - 9y' = 0 \] The characteristic equation is: \[ r^3 - 9r = 0 \implies r(r^2 - 9) = 0 \implies r = 0, 3, -3 \] So, the complementary solution is: \[ y_c(t) = C_1 + C_2e^{3t} + C_3e^{-3t} \] ### 3. **Right-hand Side (Nonhomogeneous Term):** \[ te^{-3t} + 4\cos 3t \] #### **a) For \( te^{-3t} \):** Normally, for \( e^{\alpha t} \) you try \( Ae^{\alpha t} \). But for \( te^{\alpha t} \), you try \( (At + B)e^{\alpha t} \). However, since \( e^{-3t} \) is already a root of the homogeneous equation, we need to multiply by \( t \) again to avoid duplication. So, the guess for this part is: \[ (at^2 + bt)e^{-3t} \] #### **b) For \( 4\cos 3t \):** Normally, for \( \cos(3t) \) and \( \sin(3t) \), we try \( A\cos(3t) + B\sin(3t) \). Since \( \pm 3i \) are **not** roots of the characteristic equation, we just use: \[ C\cos(3t) + D\sin(3t) \] ### 4. **Combine Both Guesses:** \[ y_p(t) = (At^2 + Bt)e^{-3t} + C\cos(3t) + D\sin(3t) \] ### 5. **Check the Options:** - The only option that matches is: > **Y(t) = t(At + B)e^{-3t} + C\cos(3t) + D\sin(3t)** This is equivalent to \( (At^2 + Bt)e^{-3t} + C\cos(3t) + D\sin(3t) \). ### **Final Answer:** \[ \boxed{\text{Y(t) = t(At + B)e^{-3t} + C\cos3t + D\sin3t}} \] So, the correct option is the **fourth one**.

Let's analyze each option carefully, considering the method of undetermined coefficients and the nature of the right-hand side of the differential equation: \[ y''' - 9 y' = t e^{-3t} + 4 \cos 3t \] --- ### Step 1: Homogeneous equation and its roots The homogeneous part: \[ y''' - 9 y' = 0 \] Characteristic equation: \[ r^3 - 9 r = 0 \implies r(r^2 - 9) = 0 \implies r = 0, \pm 3 \] **Roots:** - \( r = 0 \) - \( r = 3 \) - \( r = -3 \) **Complementary solution:** \[ y_c(t) = C_1 + C_2 e^{3t} + C_3 e^{-3t} \] --- ### Step 2: Nature of the RHS and the particular solution The RHS: \[ t e^{-3t} + 4 \cos 3t \] **Part A: \( t e^{-3t} \)** - Since \( e^{-3t} \) corresponds to a root \( r = -3 \) of the homogeneous equation, the standard trial \( A e^{-3t} \) or \( (A t + B) e^{-3t} \) **must be modified** to account for the duplication of roots. - **For \( e^{\alpha t} \) where \( \alpha \) is a root**, the trial is multiplied by \( t \) to the power equal to the multiplicity (here, multiplicity 1), so: \[ \boxed{\text{Guess: } (A t + B) e^{-3t}} \] - But since \( e^{-3t} \) corresponds to a root in the homogeneous solution, **we need to multiply further by \( t \)** to ensure linear independence, making the trial: \[ \boxed{(A t^2 + B t) e^{-3t}} \] --- **Part B: \( 4 \cos 3t \)** - Since \( r = \pm 3i \), the solutions to the homogeneous equation are \( e^{\pm 3i t} \), which correspond to \( \cos 3t \) and \( \sin 3t \). - **Standard guess for \( \cos 3t \) and \( \sin 3t \):** \[ A \cos 3t + B \sin 3t \] - **Check if these are solutions to the homogeneous equation:** The roots \( r = \pm 3i \) are **not** roots of the characteristic equation (which are \( 0, 3, -3 \)). Therefore, **no modification** is necessary, and these can be used directly. --- ### **Step 3: Construct the particular solution** Combining both parts, the **general form** of the particular solution \( y_p(t) \): \[ \boxed{ \text{Y}_p(t) = (A t^2 + B t) e^{-3t} + C \cos 3t + D \sin 3t } \] --- ### **Step 4: Options analysis** Now, let's analyze each provided option: | Option | Expression | Explanation | |----------|--------------|--------------| | **1** | \(\quad Y(t) = A e^{k t} + B t \cos 3t + C t \sin 3t\) | **Not suitable** — does not account for the duplication of \( e^{-3t} \) for \( t e^{-3t} \) and misses the quadratic \( t^2 \) term. | | **2** | \(\quad Y(t) = A P e^{k t} + B e^{3t} + C \sin 3t\) | **Incorrect notation** and terms don't match the required form. | | **3** | \(\quad Y(t) = A e + B \cos 3t + C \sin 3t\) | **Incorrect** — constant \( e \), no exponential multiplied by \( t \), and missing \( t^2 \) term for \( t e^{-3t} \). | | **4** | \(\quad Y(t) = t(At + B) e^{-3t} + C \cos 3t + D \sin 3t\) | **Correct** — matches the derived form, with the quadratic \( t^2 \) term for \( t e^{-3t} \) and the sinusoidal components. | | **5** | \(\quad Y(t) = (At + B) e^{\lambda t} + C \cos 3t + D \sin 3t\) | **Incomplete** — lacks the \( t^2 \) factor for the \( t e^{-3t} \) term. | --- ### **Final conclusion:** The **correct form** for the particular solution, suitable for the method of undetermined coefficients considering the roots and multiplicities, is: \[ \boxed{ Y(t) = t(At + B) e^{-3t} + C \cos 3t + D \sin 3t } \] which matches **Option 4**. --- ### **Final Answer:** **Option 4**: \[ \boxed{ Y(t) = t(At + B) e^{-3t} + C \cos 3t + D \sin 3t } \] This form takes into account the multiplicity of roots and the nature of the RHS functions, ensuring the method of undetermined coefficients can be correctly applied without immediately solving for the constants.

Let's analyze each option carefully and explain why it is suitable or not, based on the method of undetermined coefficients and the structure of the differential equation: Given differential equation: \[ y''' - 9 y' = t e^{-3t} + 4 \cos 3t \] --- ### Step 1: Homogeneous equation and its roots Homogeneous equation: \[ y''' - 9 y' = 0 \] Characteristic equation: \[ r^3 - 9 r = 0 \implies r(r^2 - 9) = 0 \implies r = 0, \pm 3 \] **Roots:** - \( r=0 \) - \( r=3 \) - \( r=-3 \) **Complementary (homogeneous) solution:** \[ y_c(t) = C_1 + C_2 e^{3t} + C_3 e^{-3t} \] --- ### Step 2: Analyzing the RHS and designing the particular solution The RHS: \[ t e^{-3t} + 4 \cos 3t \] --- ### Part A: \( t e^{-3t} \) - The exponential part, \( e^{-3t} \), has root \( r = -3 \) in the characteristic roots. - For \( e^{\alpha t} \) where \( \alpha \) is a root, the usual guess would be \( A e^{\alpha t} \). - For \( t e^{\alpha t} \), the guess is \( (A t + B) e^{\alpha t} \). - **However**, since \( e^{-3t} \) corresponds to a root \( r = -3 \) (already in the homogeneous solution), we need to multiply by \( t \) **again** to get a linearly independent particular solution. This leads to: \[ (at^2 + bt) e^{-3t} \] ### Part B: \( 4 \cos 3t \) - Roots \( r = \pm 3 i \) correspond to \( \cos 3t \) and \( \sin 3t \). - Since these are **not** roots of the homogeneous characteristic equation, the standard guess is: \[ A \cos 3t + B \sin 3t \] --- ### **Step 3: Constructing the particular solution** Combining both parts, the suitable particular solution form is: \[ Y_p(t) = (A t^2 + B t) e^{-3t} + C \cos 3t + D \sin 3t \] --- ### **Step 4: Examining the options** Now, let's see which option matches this form: | Option | Expression | Explanation | |---------|--------------|--------------| | **1** | \( Y(t) = A e^{k t} + B t \cos 3t + C t \sin 3t \) | **Incorrect**. Does not include the \( e^{-3t} \) exponential nor the quadratic \( t^2 \) term needed for \( t e^{-3t} \). | | **2** | \( Y(t) = A P e^{k t} + B e^{3t} + C \sin 3t \) | **Incorrect and unclear notation**. Does not match the required form. | | **3** | \( Y(t) = A e + B \cos 3t + C \sin 3t \) | **Incorrect**. Constant \( e \), no exponential part with \( t e^{-3t} \), and missing the \( t^2 \) term. | | **4** | \( Y(t) = t(At + B) e^{-3t} + C \cos 3t + D \sin 3t \) | **Correct**. Matches the derived form, with the quadratic \( t^2 \) factor for \( t e^{-3t} \) and sinusoidal parts. | | **5** | \( Y(t) = (At + B) e^{\lambda t} + C \cos 3t + D \sin 3t \) | **Incomplete**. Lacks the \( t^2 \) factor necessary for \( t e^{-3t} \). | --- ### **Final conclusion:** The **correct form** of the particular solution, considering roots and multiplicities, is: \[ \boxed{ Y(t) = t(At + B) e^{-3t} + C \cos 3t + D \sin 3t } \] which is exactly **Option 4**. --- ### **Summary:** - **Option 1:** Fails to account for root multiplicity and the specific \( t e^{-3t} \) term. - **Option 2:** Not aligned with the required form. - **Option 3:** Doesn't include the exponential \( e^{-3t} \) for the \( t e^{-3t} \) term. - **Option 4:** Correctly incorporates the \( t^2 e^{-3t} \) structure for \( t e^{-3t} \) and sinusoidal components. - **Option 5:** Missing the quadratic \( t^2 \) factor, so not suitable. --- ### **Final answer:** \[ \boxed{ \text{Option 4: } \quad Y(t) = t(At + B) e^{-3t} + C \cos 3t + D \sin 3t } \] This form ensures the method of undetermined coefficients can be correctly applied without conflicts with the roots of the homogeneous equation.

Let's analyze each option carefully, understanding their structure and appropriateness for the given differential equation: --- ## Given Differential Equation: \[ y''' - 9 y' = t e^{-3t} + 4 \cos 3t \] --- ## Step 1: Homogeneous Equation and Roots Homogeneous form: \[ y''' - 9 y' = 0 \] Characteristic equation: \[ r^3 - 9 r = 0 \implies r(r^2 - 9) = 0 \implies r=0, \pm 3 \] **Roots:** - \( r=0 \) - \( r=3 \) - \( r=-3 \) **Complementary solution:** \[ y_c(t) = C_1 + C_2 e^{3t} + C_3 e^{-3t} \] --- ## Step 2: Analyzing the RHS and suitable particular solution forms RHS: \[ t e^{-3t} + 4 \cos 3t \] ### Part A: \( t e^{-3t} \) - Since \( e^{-3t} \) corresponds to the root \( r = -3 \), which appears in the homogeneous solution, the usual guess \( A e^{-3t} \) is not sufficient. - For \( t e^{\alpha t} \) where \( \alpha \) is a root, the trial is multiplied by \( t \). But because the root \( r = -3 \) is already in the homogeneous solution, to get a linearly independent particular solution, we need to multiply by \( t \) **again**, resulting in: \[ \boxed{(A t^2 + B t) e^{-3t}} \] ### Part B: \( 4 \cos 3t \) - Roots \( r= \pm 3i \) relate to \( \cos 3t \) and \( \sin 3t \). - Since these are **not** roots of the characteristic equation, the standard guess is: \[ A \cos 3t + B \sin 3t \] --- ## Step 3: General form of particular solution Combining both parts, the particular solution form is: \[ Y_p(t) = (A t^2 + B t) e^{-3t} + C \cos 3t + D \sin 3t \] --- ## Step 4: Examine each option ### **Option 1:** \[ Y(t) = A e^{k t} + B t \cos 3t + C t \sin 3t \] **Analysis:** - Does not include the exponential \( e^{-3t} \) specific to the RHS \( t e^{-3t} \). - Also, the sinusoidal parts are multiplied by \( t \), which is unnecessary. - **Incorrect** for the given RHS. --- ### **Option 2:** \[ Y(t) = A P e^{k t} + B e^{3t} + C \sin 3t \] **Analysis:** - Notation \( P e^{k t} \) is unclear. - Includes \( e^{3t} \) (which is part of the homogeneous solution and may require modification if used as a particular solution). - Omits the \( e^{-3t} \) term altogether. - The sinusoid \( C \sin 3t \) is present, but the form is incomplete. - **Incorrect**. --- ### **Option 3:** \[ Y(t) = A e + B \cos 3t + C \sin 3t \] **Analysis:** - Contains a constant \( e \) (which is unrelated). - No exponential component with \( e^{-3t} \) or \( t e^{-3t} \). - Fails to account for the multiplicity and the polynomial factor for \( t e^{-3t} \). - **Incorrect**. --- ### **Option 4:** \[ Y(t) = t(At + B) e^{-3t} + C \cos 3t + D \sin 3t \] **Analysis:** - Correctly includes \( t^2 e^{-3t} \) (since \( t(At + B) = A t^2 + B t \)), suitable for \( t e^{-3t} \) where \( e^{-3t} \) is associated with a root in the characteristic equation. - Includes sinusoidal parts \( C \cos 3t + D \sin 3t \). - Matches the derived form for the particular solution. - **Correct**. --- ### **Option 5:** \[ Y(t) = (A t + B) e^{\lambda t} + C \cos 3t + D \sin 3t \] **Analysis:** - Lacks the \( t^2 \) factor necessary for \( t e^{-3t} \). - Also, \( e^{\lambda t} \) is ambiguous; it should specify \( e^{-3t} \) to be consistent. - **Incorrect**. --- ## **Final Conclusion:** The most appropriate form, considering the roots and multiplicities, is: \[ \boxed{ Y(t) = t(At + B) e^{-3t} + C \cos 3t + D \sin 3t } \] which **matches Option 4**. --- ## **Summary:** | **Option** | **Reasoning** | **Correct/Incorrect** | |--------------|----------------|------------------------| | **1** | Does not include the necessary \( e^{-3t} \) and polynomial factors | Incorrect | | **2** | Unclear notation, missing key exponential term | Incorrect | | **3** | Lacks exponential \( e^{-3t} \), no polynomial for \( t e^{-3t} \) | Incorrect | | **4** | Correctly includes \( t^2 e^{-3t} \) and sinusoidal parts | **Correct** | | **5** | Missing the \( t^2 \) factor, incomplete form | Incorrect | --- ## **Final Answer:** \[ \boxed{ \textbf{Option 4:} \quad Y(t) = t(At + B) e^{-3t} + C \cos 3t + D \sin 3t } \] This form correctly accommodates the multiplicity of roots and the structure of the RHS, enabling the method of undetermined coefficients to be applied effectively.