Let's solve **(a)** step by step using the **D-operator method** and the **exponential-shift rule**: ### (a) \(\frac{d^2y}{dx^2} + 5\frac{dy}{dx} - 6y = 3e^{-x} + x, \quad y(0) = 1, \quad y'(0) = 0\) --- #### **Step 1: Rewrite Using the D-operator** Let \( D = \frac{d}{dx} \): \[ D^2y + 5Dy - 6y = 3e^{-x} + x \] \[ (D^2 + 5D - 6)y = 3e^{-x} + x \] --- #### **Step 2: Solve the Homogeneous Equation** \[ (D^2 + 5D - 6)y = 0 \] Find the characteristic equation: \[ D^2 + 5D - 6 = 0 \] \[ (D + 6)(D - 1) = 0 \implies D = -6, 1 \] So, the complementary solution is: \[ y_c = C_1 e^{-6x} + C_2 e^{x} \] --- #### **Step 3: Find a Particular Solution \(y_p\)** The nonhomogeneous part is \(3e^{-x} + x\). Find a particular solution for each term: ##### **(i) For \(3e^{-x}\):** Apply the exponential-shift rule: \[ (D^2 + 5D - 6)(Ae^{-x}) = 3e^{-x} \] Let \(y_1 = Ae^{-x}\): Using shift rule: \(D \to D - 1\) when applied to \(e^{-x}\): \[ (D^2 + 5D - 6)y_1 = (D^2 + 5D - 6)Ae^{-x} = e^{-x}[(D-(-1))^2 + 5(D-(-1)) - 6]A = e^{-x}[(D+1)^2 + 5(D+1) - 6]A \] But it is simpler to substitute \(y_1 = Ae^{-x}\) directly: \[ y_1' = -Ae^{-x},\quad y_1'' = Ae^{-x} \] \[ y_1'' + 5y_1' - 6y_1 = Ae^{-x} - 5Ae^{-x} - 6Ae^{-x} = (A - 5A - 6A)e^{-x} = (-10A)e^{-x} \] Set equal to \(3e^{-x}\): \[ -10A = 3 \implies A = -\frac{3}{10} \] So, \[ y_1 = -\frac{3}{10}e^{-x} \] --- ##### **(ii) For \(x\):** Try \(y_2 = Bx + C\): \[ y_2' = B,\quad y_2'' = 0 \] Plug into the equation: \[ 0 + 5B - 6(Bx + C) = x \] \[ 5B - 6Bx - 6C = x \] Match coefficients: - Coefficient of \(x\): \(-6B = 1 \implies B = -\frac{1}{6}\) - Constant: \(5B - 6C = 0 \implies 5(-\frac{1}{6}) - 6C = 0 \implies -\frac{5}{6} - 6C = 0 \implies C = -\frac{5}{36}\) So, \[ y_2 = -\frac{1}{6}x - \frac{5}{36} \] --- #### **Step 4: General Solution** \[ y(x) = y_c + y_p = C_1 e^{-6x} + C_2 e^{x} - \frac{3}{10}e^{-x} - \frac{1}{6}x - \frac{5}{36} \] --- #### **Step 5: Apply Initial Conditions** Given: \(y(0) = 1, \quad y'(0) = 0\) Find \(y(0):\) \[ y(0) = C_1 + C_2 - \frac{3}{10} - 0 - \frac{5}{36} = 1 \] Find \(y'(x):\) \[ y'(x) = -6C_1 e^{-6x} + C_2 e^{x} + \frac{3}{10}e^{-x} - \frac{1}{6} \] So, \[ y'(0) = -6C_1 + C_2 + \frac{3}{10} - \frac{1}{6} = 0 \] Set up the system: \[ \begin{cases} C_1 + C_2 - \frac{3}{10} - \frac{5}{36} = 1 \\ -6C_1 + C_2 + \frac{3}{10} - \frac{1}{6} = 0 \end{cases} \] Compute \(-\frac{3}{10} - \frac{5}{36}\): LCD is 180: \[ -\frac{3}{10} = -\frac{54}{180}, \quad -\frac{5}{36} = -\frac{25}{180} \] \[ -\frac{54}{180} - \frac{25}{180} = -\frac{79}{180} \] So, \[ C_1 + C_2 - \frac{79}{180} = 1 \implies C_1 + C_2 = 1 + \frac{79}{180} = \frac{259}{180} \] Now, for the second equation: \[ \frac{3}{10} - \frac{1}{6} = \frac{9-5}{30} = \frac{4}{30} = \frac{2}{15} \] \[ -6C_1 + C_2 + \frac{2}{15} = 0 \implies -6C_1 + C_2 = -\frac{2}{15} \] Now solve the system: \[ \begin{cases} C_1 + C_2 = \frac{259}{180} \\ -6C_1 + C_2 = -\frac{2}{15} \end{cases} \] Subtract equations: Let’s solve for \(C_1\): Subtract the first from the second: \[ (-6C_1 + C_2) - (C_1 + C_2) = -\frac{2}{15} - \frac{259}{180} \] \[ -7C_1 = -\frac{2}{15} - \frac{259}{180} \] Convert \(-\frac{2}{15}\) to denominator 180: \[ -\frac{2}{15} = -\frac{24}{180} \] \[ -7C_1 = -\frac{24}{180} - \frac{259}{180} = -\frac{283}{180} \] \[ C_1 = \frac{283}{1260} \] Now, \(C_2 = \frac{259}{180} - \frac{283}{1260}\) Convert both to denominator 1260: \[ \frac{259}{180} = \frac{259 \times 7}{1260} = \frac{1813}{1260} \] \[ C_2 = \frac{1813 - 283}{1260} = \frac{1530}{1260} = \frac{17}{14} \] --- #### **Step 6: Final Solution** \[ \boxed{ y(x) = \frac{283}{1260}e^{-6x} + \frac{17}{14}e^{x} - \frac{3}{10}e^{-x} - \frac{1}{6}x - \frac{5}{36} } \] --- Would you like the solutions for parts **(b)–(f)** as well?

Certainly! Let's proceed step-by-step with parts (b) to (f) using the D-operator method and the exponential-shift rule. --- ### **(b)** Solve \(6D y = 3t e^{2t}\) --- #### **Step 1: Rewrite in operator form** \[ 6D y = 3t e^{2t} \] \[ D y = \frac{1}{2} t e^{2t} \] --- #### **Step 2: Find \(y(t)\)** Since \(D y = \frac{1}{2} t e^{2t}\), integrate both sides: \[ y(t) = \int \frac{1}{2} t e^{2t} dt + C \] Use integration by parts: Let: - \(u = t \Rightarrow du = dt\) - \(dv = e^{2t} dt \Rightarrow v = \frac{1}{2} e^{2t}\) Then: \[ \int t e^{2t} dt = t \cdot \frac{1}{2} e^{2t} - \int \frac{1}{2} e^{2t} dt = \frac{t e^{2t}}{2} - \frac{1}{4} e^{2t} + C \] Therefore, \[ y(t) = \frac{1}{2} \left( \frac{t e^{2t}}{2} - \frac{1}{4} e^{2t} \right) + C = \frac{t e^{2t}}{4} - \frac{1}{8} e^{2t} + C \] --- ### **(c)** Solve \(D y - 2 y = \sin 2 t\) --- #### **Step 1: Operator form** \[ (D - 2) y = \sin 2 t \] --- #### **Step 2: Find the particular solution** Apply the exponential shift rule: \[ (D - 2) y = \sin 2 t \] Recall that \(\sin 2 t\) corresponds to the operator \(D\): \[ (D - 2) y = \sin 2 t \] Using the inverse operator: \[ y = \frac{1}{D - 2} \sin 2 t \] Express \(\sin 2 t\) as the imaginary part of \(e^{i 2 t}\): \[ \sin 2 t = \Im (e^{i 2 t}) \] Solve: \[ (D - 2) y = e^{i 2 t} \] The *exponential-shift rule* states: \[ \frac{1}{D - a} f(t) = e^{a t} \int e^{-a t} f(t) dt \] Applying this: \[ y(t) = e^{2 t} \int e^{-2 t} e^{i 2 t} dt = e^{2 t} \int e^{(-2 + i 2) t} dt \] \[ = e^{2 t} \int e^{2 t (i - 1)} dt \] Evaluate: \[ \int e^{2 t (i - 1)} dt = \frac{1}{2(i - 1)} e^{2 t (i -1)} + C \] Now, \[ y(t) = e^{2 t} \cdot \frac{1}{2(i - 1)} e^{2 t (i - 1)} = \frac{1}{2(i - 1)} e^{2 t} e^{2 t (i - 1)} \] \[ = \frac{1}{2(i - 1)} e^{2 t (1 + i - 1)} = \frac{1}{2(i - 1)} e^{2 i t} \] Similarly, the particular solution for the real part: \[ y(t) = \Re \left( \frac{1}{2(i - 1)} e^{2 i t} \right) \] Compute denominator: \[ i - 1 = -1 + i \] Its reciprocal: \[ \frac{1}{-1 + i} = \frac{-1 - i}{(-1)^2 + 1^2} = \frac{-1 - i}{2} \] Thus, \[ \frac{1}{2(i - 1)} = \frac{1}{2} \times \frac{-1 - i}{2} = \frac{-1 - i}{4} \] Multiply by \(e^{2 i t}\): \[ y(t) = \Re \left( \frac{-1 - i}{4} e^{2 i t} \right) \] Express \(e^{2 i t} = \cos 2 t + i \sin 2 t\): \[ y(t) = \Re \left[ \frac{-1 - i}{4} (\cos 2 t + i \sin 2 t) \right] \] Multiply out numerator: \[ \frac{-1 - i}{4} (\cos 2 t + i \sin 2 t) = \frac{1}{4} [(-1)(\cos 2 t + i \sin 2 t) + (-i)(\cos 2 t + i \sin 2 t)] \] Compute: \[ = \frac{1}{4} \left( -\cos 2 t - i \sin 2 t - i \cos 2 t - i^2 \sin 2 t \right) \] Recall \(i^2 = -1\): \[ = \frac{1}{4} \left( -\cos 2 t - i \sin 2 t - i \cos 2 t + \sin 2 t \right) \] Group real and imaginary parts: Real: \[ -\cos 2 t + \sin 2 t \] Imaginary: \[ - i \sin 2 t - i \cos 2 t = - i (\sin 2 t + \cos 2 t) \] Since we only need the real part: \[ \boxed{ y(t) = \frac{1}{4} \left( - \cos 2 t + \sin 2 t \right) } \] --- ### **(d)** Solve \(w'(t) + 6 w(t) = 0\) This is a simple first-order homogeneous ODE: \[ w'(t) = -6 w(t) \] Solution: \[ w(t) = C e^{-6 t} \] --- ### **(e)** Solve \(D y - 8 y = 0\) Same as (c), with \(a = 8\): \[ (D - 8) y = 0 \] \[ \Rightarrow y(t) = C e^{8 t} \] --- ### **(f)** Solve \(D^2 y + 8 y = 0\) Characteristic equation: \[ D^2 + 8 = 0 \implies D^2 = -8 \] \[ D = \pm i \sqrt{8} = \pm i 2 \sqrt{2} \] Solution: \[ \boxed{ y(t) = A \cos(2 \sqrt{2} t) + B \sin(2 \sqrt{2} t) } \] --- ## **Summary of solutions:** | Part | Equation | Solution | |---------|------------------------------|-------------------------------------------------------------| | (a) | \(\frac{d^2 y}{dx^2} + 5 \frac{dy}{dx} - 6 y = 3 e^{-x} + x\) | \( y = C_1 e^{-6x} + C_2 e^{x} - \frac{3}{10} e^{-x} - \frac{1}{6} x - \frac{5}{36} \) | | (b) | \(6 \frac{dy}{dt} = 3 t e^{2 t}\) | \( y(t) = \frac{t e^{2 t}}{4} - \frac{1}{8} e^{2 t} + C \) | | (c) | \(\frac{dy}{dt} - 2 y = \sin 2 t\) | \( y(t) = \frac{- \cos 2 t + \sin 2 t}{4} \) | | (d) | \(w'(t) + 6 w(t) = 0\) | \( w(t) = C e^{-6 t} \) | | (e) | \(D y - 8 y = 0\) | \( y(t) = C e^{8 t} \) | | (f) | \(D^2 y + 8 y = 0\) | \( y(t) = A \cos(2 \sqrt{2} t) + B \sin(2 \sqrt{2} t) \) | --- Would you like further clarification on any step?