Let's solve each differential equation using the **method of undetermined coefficients**. --- ## **Problem 9** **Equation:** \( 2y'' - 3y' = 9x^2 - 10 \), with \( y(0) = 1 \), \( y'(0) = -3 \). ### **Step 1: Solve the homogeneous equation** \( 2y'' - 3y' = 0 \) The characteristic equation: \[ 2r^2 - 3r = 0 \implies r(2r - 3) = 0 \implies r = 0,\, \frac{3}{2} \] So the homogeneous solution: \[ y_h(x) = C_1 + C_2 e^{\frac{3}{2}x} \] --- ### **Step 2: Find a particular solution** Guess a particular solution of the form: \[ y_p = ax^2 + bx + c \] Compute derivatives: - \( y_p' = 2ax + b \) - \( y_p'' = 2a \) Plug into the differential equation: \[ 2y'' - 3y' = 2(2a) - 3(2ax + b) = 4a - 6ax - 3b = 9x^2 - 10 \] But notice that the right side is \(9x^2 - 10\), so our guess must include an \(x^2\) term. But since the left side, as written, will never generate an \(x^2\) from such a guess, let's try: \[ y_p = Ax^3 + Bx^2 + Cx + D \] But since the right side is quadratic, let's check: - \(y_p = Ax^2 + Bx + C\) Compute: - \( y_p' = 2Ax + B \) - \( y_p'' = 2A \) Plug into the equation: \[ 2y_p'' - 3y_p' = 2(2A) - 3(2Ax + B) = 4A - 6Ax - 3B \] Set equal to \(9x^2 - 10\): \[ 4A - 6Ax - 3B = 9x^2 - 10 \] Match coefficients: - \(x^2:\)         0 = 9        → Contradiction This means our guess is missing the degree for the right side. Let's guess: \[ y_p = Px^3 + Qx^2 + Rx + S \] Compute: - \(y_p' = 3Px^2 + 2Qx + R\) - \(y_p'' = 6Px + 2Q\) Plug into the equation: \[ 2y_p'' - 3y_p' = 2(6Px + 2Q) - 3(3Px^2 + 2Qx + R) = 12Px + 4Q - 9Px^2 - 6Qx - 3R = (-9P)x^2 + (12P - 6Q)x + (4Q - 3R) \] Set equal to right side: \[ -9Px^2 + (12P - 6Q)x + (4Q - 3R) = 9x^2 - 10 \] Match coefficients: - \(x^2:\)  \(-9P = 9 \implies P = -1\) - \(x^1:\)  \(12P - 6Q = 0 \implies 12(-1) - 6Q = 0 \implies -12 - 6Q = 0 \implies Q = -2\) - Constant: \(4Q - 3R = -10 \implies 4(-2) - 3R = -10 \implies -8 - 3R = -10 \implies -3R = -2 \implies R = \frac{2}{3}\) So, \[ y_p = -x^3 - 2x^2 + \frac{2}{3}x \] --- ### **Step 3: General solution** \[ y(x) = C_1 + C_2 e^{\frac{3}{2}x} - x^3 - 2x^2 + \frac{2}{3}x \] --- ### **Step 4: Apply initial conditions** **\( y(0) = 1 \):** \[ 1 = C_1 + C_2 - 0 - 0 + 0 \implies C_1 + C_2 = 1 \] **\( y'(x) = \)** \[ y'(x) = C_2 \cdot \frac{3}{2} e^{\frac{3}{2}x} - 3x^2 - 4x + \frac{2}{3} \] So at \(x=0\): \[ y'(0) = C_2 \cdot \frac{3}{2} + 0 + 0 + \frac{2}{3} = -3 \implies \frac{3}{2}C_2 + \frac{2}{3} = -3 \implies \frac{3}{2}C_2 = -3 - \frac{2}{3} = -\frac{11}{3} \implies C_2 = -\frac{11}{3} \cdot \frac{2}{3} = -\frac{22}{9} \] Then, \(C_1 = 1 - C_2 = 1 + \frac{22}{9} = \frac{31}{9}\) --- ### **Final Answer for 9:** \[ \boxed{ y(x) = \frac{31}{9} - \frac{22}{9} e^{\frac{3}{2}x} - x^3 - 2x^2 + \frac{2}{3}x } \] --- ## **Problem 10** **Equation:** \[ 3y'' - 5y' - 2y = 3e^{2t} \] ### **Step 1: Homogeneous Solution** Characteristic equation: \[ 3r^2 - 5r - 2 = 0 \] Solve: \[ r = \frac{5 \pm \sqrt{25 + 24}}{6} = \frac{5 \pm 7}{6} \implies r_1 = 2, \quad r_2 = -\frac{1}{3} \] So, \[ y_h(t) = C_1 e^{2t} + C_2 e^{-\frac{1}{3}t} \] --- ### **Step 2: Particular Solution** Right side is \(3e^{2t}\), but \(e^{2t}\) is already a solution to the homogeneous equation, so **try**: \[ y_p = At e^{2t} \] Compute derivatives: \[ y_p' = A e^{2t} + 2A t e^{2t} \] \[ y_p'' = 2A e^{2t} + 2A e^{2t} + 4A t e^{2t} = 4A e^{2t} + 4A t e^{2t} \] Plug into the equation: \[ 3y_p'' - 5y_p' - 2y_p = 3[4A e^{2t} + 4A t e^{2t}] - 5[A e^{2t} + 2A t e^{2t}] - 2 (A t e^{2t}) \] \[ = [12A e^{2t} + 12A t e^{2t}] - [5A e^{2t} + 10A t e^{2t}] - [2A t e^{2t}] \] \[ = (12A - 5A) e^{2t} + (12A - 10A - 2A) t e^{2t} = 7A e^{2t} + 0 \cdot t e^{2t} \] Set equal to \(3e^{2t}\): \[ 7A e^{2t} = 3e^{2t} \implies A = \frac{3}{7} \] So, particular solution: \[ y_p = \frac{3}{7} t e^{2t} \] --- ### **General Solution:** \[ \boxed{ y(t) = C_1 e^{2t} + C_2 e^{-\frac{1}{3}t} + \frac{3}{7} t e^{2t} } \] --- If you need step-by-step explanations or more details, let me know!