Let's solve **Question 1.1** step-by-step using the **D-operator** method. --- ## **Question 1.1** Given: \[ (D^2 - 2D + 5)y = x + 5 \] where \( D = \frac{d}{dx} \). --- ### **Step 1: Solve the Homogeneous Equation** \[ (D^2 - 2D + 5)y = 0 \] The auxiliary equation is: \[ m^2 - 2m + 5 = 0 \] Solve for \( m \): \[ m = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 5}}{2} = \frac{2 \pm \sqrt{4 - 20}}{2} = \frac{2 \pm \sqrt{-16}}{2} = \frac{2 \pm 4i}{2} = 1 \pm 2i \] So, the complementary function (CF) is: \[ y_{CF} = e^{x}(A\cos2x + B\sin2x) \] where \(A\) and \(B\) are arbitrary constants. --- ### **Step 2: Find a Particular Integral (PI)** \[ y_{PI} = \frac{1}{D^2 - 2D + 5}(x + 5) \] Let’s solve for PI for each term separately. #### **For the '5' term:** \[ y_{PI, 1} = \frac{1}{D^2 - 2D + 5}(5) \] Since 5 is a constant, try \(y = K\): \[ (D^2 - 2D + 5)K = 5 \implies 5K = 5 \implies K = 1 \] #### **For the 'x' term:** Try \(y = ax + b\): \[ (D^2 - 2D + 5)(ax + b) = x \] First, compute derivatives: - \(y = ax + b\) - \(y' = a\) - \(y'' = 0\) Plug these in: \[ 0 - 2(a) + 5(ax + b) = x \] \[ -2a + 5a x + 5b = x \] \[ 5a x + (5b - 2a) = x \] Equate coefficients: - Coefficient of \(x\): \(5a = 1 \implies a = \frac{1}{5}\) - Constant term: \(5b - 2a = 0 \implies b = \frac{2a}{5} = \frac{2}{25}\) So, the PI for \(x\) is \(y_{PI,2} = \frac{1}{5}x + \frac{2}{25}\). #### **Sum the particular integrals:** Total PI: \[ y_{PI} = 1 + \frac{1}{5}x + \frac{2}{25} = \frac{1}{5}x + \frac{27}{25} \] --- ### **Step 3: General Solution** \[ \boxed{ y = e^{x}(A\cos2x + B\sin2x) + \frac{1}{5}x + \frac{27}{25} } \] --- #### **Summary** - **Complementary Function (CF):** \( e^{x}(A\cos2x + B\sin2x) \) - **Particular Integral (PI):** \( \frac{1}{5}x + \frac{27}{25} \) - **General Solution:** \[ \boxed{ y = e^{x}(A\cos2x + B\sin2x) + \frac{1}{5}x + \frac{27}{25} } \] --- If you want step-by-step solutions for 1.2, 1.3, or the other questions, please let me know!

Let's systematically solve all parts of the questions with detailed steps. --- ## **Question 1: Find the general solutions of the differential equations using D-operator methods** --- ### **Part 1.1: \((D^2 - 2D + 5)y = x + 5\)** **Step 1: Homogeneous solution** Given: \[ (D^2 - 2D + 5) y = 0 \] **Auxiliary equation:** \[ m^2 - 2m + 5 = 0 \] \[ m = \frac{2 \pm \sqrt{(-2)^2 - 4 \times 1 \times 5}}{2} = \frac{2 \pm \sqrt{4 - 20}}{2} = \frac{2 \pm \sqrt{-16}}{2} = \frac{2 \pm 4i}{2} = 1 \pm 2i \] **Homogeneous solution:** \[ y_h = e^{x} (A \cos 2x + B \sin 2x) \] --- **Step 2: Find particular integral (PI)** - RHS is \(x + 5\), a polynomial in \(x\). For polynomial RHS, assume PI as polynomial form. - Since the operator is second order with complex roots, and RHS is polynomial, try: \[ y_p = P x + Q \] - Apply the operator to \(y_p\): \[ D y_p = P \] \[ D^2 y_p = 0 \] **Compute:** \[ (D^2 - 2D + 5) y_p = 0 - 2 P + 5 (P x + Q) \] \[ = -2 P + 5 P x + 5 Q \] Set equal to RHS: \[ -2 P + 5 P x + 5 Q = x + 5 \] Matching coefficients: - Coefficient of \(x\): \[ 5 P = 1 \Rightarrow P = \frac{1}{5} \] - Constant term: \[ -2 P + 5 Q = 5 \] \[ -2 \times \frac{1}{5} + 5 Q = 5 \] \[ -\frac{2}{5} + 5 Q = 5 \] \[ 5 Q = 5 + \frac{2}{5} = \frac{25}{5} + \frac{2}{5} = \frac{27}{5} \] \[ Q = \frac{27}{25} \] **Particular solution:** \[ y_p = \frac{1}{5} x + \frac{27}{25} \] --- ### **Final solution:** \[ \boxed{ \boxed{ y = e^{x} (A \cos 2x + B \sin 2x) + \frac{1}{5} x + \frac{27}{25} } } \] --- ## **Part 1.2: \((D^2 + 2D + 5) y = e^{z} \cos 3z\)** **Note:** The variable is \(z\); the operator is with respect to \(z\). ### **Step 1: Homogeneous solution** Auxiliary equation: \[ m^2 + 2 m + 5 = 0 \] \[ m = \frac{-2 \pm \sqrt{4 - 20}}{2} = \frac{-2 \pm \sqrt{-16}}{2} = -1 \pm 2i \] **Homogeneous solution:** \[ y_h = e^{-z} (C \cos 2z + D \sin 2z) \] ### **Step 2: Find particular integral** - RHS: \(e^{z} \cos 3z\) - To handle RHS, note that the operator \(D^2 + 2D + 5\) acts on functions of \(z\). - Since RHS involves \(e^{z} \cos 3z\), try a particular solution of the form: \[ y_p = e^{z} (A \cos 3z + B \sin 3z) \] --- ### **Step 3: Apply the operator** Recall: \[ (D - r)^n e^{r z} \cos \omega z \text{ or } e^{r z} \sin \omega z \] The operator: \[ D^2 + 2D + 5 \] acts on \(y_p\). Let's compute: \[ D y_p = \frac{d}{dz} \left[ e^{z} (A \cos 3z + B \sin 3z) \right] \] Using product rule: \[ D y_p = e^{z} (A \cos 3z + B \sin 3z) + e^{z} \left[ -3A \sin 3z + 3B \cos 3z \right] \] \[ = e^{z} \left[ (A \cos 3z + B \sin 3z) + (-3A \sin 3z + 3 B \cos 3z) \right] \] \[ = e^{z} \left[ (A + 3 B) \cos 3z + (B - 3 A) \sin 3z \right] \] Similarly, \[ D^2 y_p = D (D y_p) \] Differentiate \(D y_p\): \[ D^2 y_p = \frac{d}{dz} \left[ e^{z} \left( (A + 3 B) \cos 3z + (B - 3A) \sin 3z \right) \right] \] Again, product rule: \[ = e^{z} \left( (A + 3 B) \cos 3z + (B - 3A) \sin 3z \right) + e^{z} \left[ -3 (A + 3 B) \sin 3z + 3 (B - 3A) \cos 3z \right] \] Simplify: \[ D^2 y_p = e^{z} \left[ (A + 3 B) \cos 3z + (B - 3A) \sin 3z - 3 (A + 3 B) \sin 3z + 3 (B - 3A) \cos 3z \right] \] Group \(\cos 3z\) and \(\sin 3z\): \[ = e^{z} \left[ \left( (A + 3 B) + 3 (B - 3A) \right) \cos 3z + \left( (B - 3A) - 3 (A + 3 B) \right) \sin 3z \right] \] Compute coefficients: - For \(\cos 3z\): \[ A + 3 B + 3 B - 9A = (A - 9A) + (3 B + 3 B) = -8A + 6 B \] - For \(\sin 3z\): \[ B - 3A - 3A - 9 B = (B - 9 B) - 6A = -8 B - 6A \] Thus, \[ D^2 y_p = e^{z} \left[ (-8A + 6 B) \cos 3z + (-8 B - 6 A) \sin 3z \right] \] Now, compute \( (D^2 + 2D + 5) y_p \): \[ = D^2 y_p + 2 D y_p + 5 y_p \] Recall: \[ D y_p = e^{z} [ (A + 3 B) \cos 3z + (B - 3A) \sin 3z] \] \[ 2 D y_p = 2 e^{z} [ (A + 3 B) \cos 3z + (B - 3A) \sin 3z] \] \[ 5 y_p = 5 e^{z} [A \cos 3z + B \sin 3z] \] Adding all: \[ (D^2 + 2D + 5) y_p = e^{z} \left[ (-8A + 6 B) \cos 3z + (-8 B - 6 A) \sin 3z \right] \] \[ + 2 e^{z} [ (A + 3 B) \cos 3z + (B - 3A) \sin 3z ] \] \[ + 5 e^{z} [A \cos 3z + B \sin 3z] \] Combine coefficients: \[ e^{z} \left\{ [ (-8A + 6 B) + 2(A + 3 B) + 5A ] \cos 3z + [ (-8 B - 6A) + 2(B - 3A) + 5 B ] \sin 3z \right\} \] Simplify each: **\(\cos 3z\):** \[ -8A + 6 B + 2A + 6 B + 5A = (-8A + 2A + 5A) + (6 B + 6 B) = (-8A + 7A) + 12 B = -A + 12 B \] **\(\sin 3z\):** \[ -8 B - 6A + 2 B - 6A + 5 B = (-6A - 6A) + (-8 B + 2 B + 5 B) = -12 A + (-8 B + 7 B) = -12 A - B \] **Final expression:** \[ (D^2 + 2D + 5) y_p = e^{z} [(-A + 12 B) \cos 3z + (-12 A - B) \sin 3z] \] Set equal to RHS: \[ e^{z} \cos 3z \] which implies: \[ e^{z} [(-A + 12 B) \cos 3z + (-12 A - B) \sin 3z] = e^{z} \cos 3z \] Dividing both sides by \(e^{z}\): \[ (-A + 12 B) \cos 3z + (-12 A - B) \sin 3z = \cos 3z \] Matching coefficients: \[ - A + 12 B = 1 \quad (1) \] \[ -12 A - B = 0 \quad (2) \] From (2): \[ -12 A - B = 0 \Rightarrow B = -12 A \] Substitute into (1): \[ - A + 12(-12 A) = 1 \] \[ - A - 144 A = 1 \] \[ -145 A = 1 \] \[ A = - \frac{1}{145} \] Now, \(B\): \[ B = -12 A = -12 \times \left( - \frac{1}{145} \right) = \frac{12}{145} \] **Particular solution:** \[ y_p = e^{z} \left( - \frac{1}{145} \cos 3z + \frac{12}{145} \sin 3z \right) \] --- ### **Final general solution:** \[ \boxed{ \boxed{ y(z) = e^{-z} (C \cos 2z + D \sin 2z) + e^{z} \left( - \frac{1}{145} \cos 3z + \frac{12}{145} \sin 3z \right) } } \] --- ## **Question 2: Solve for \(x\) only in the set of simultaneous equations** Given: \[ D_z y - (D - 1) y = 2z \quad \text{and} \quad z + (D - 1) y = e^z \] But the question appears to be: \[ Dz - (D - 1) y = 2z \quad \text{and} \quad z + (D - 1) y = e^z \] Assuming the equations are: \[ Dz - (D - 1) y = 2z \quad (1) \] \[ z + (D - 1) y = e^{z} \quad (2) \] --- ### **Step 1: Express \( (D - 1) y \) from equation (2)** From (2): \[ z + (D - 1) y = e^{z} \Rightarrow (D - 1) y = e^{z} - z \] ### **Step 2: Substitute into equation (1)** Equation (1): \[ Dz - (D - 1) y = 2z \] Rewrite: \[ Dz - (e^{z} - z) = 2z \] \[ Dz = 2z + e^{z} - z = z + e^{z} \] But \(Dz = \frac{d z}{d z} = 1\) (since \(z\) is independent variable, and we are solving for \(x\), but the variables are confusing