Let's solve each differential equation using the **method of undetermined coefficients**. --- ### **6. \( 2y'' + y' = 4x \), \( y(0) = 7 \), \( y'(0) = -10 \)** #### **Step 1: Solve the homogeneous equation** \( 2y'' + y' = 0 \) Characteristic equation: \[ 2r^2 + r = 0 \implies r(2r + 1) = 0 \implies r = 0,\, r = -\frac{1}{2} \] So the general solution of the homogeneous part: \[ y_h(x) = C_1 + C_2 e^{-x/2} \] #### **Step 2: Find a particular solution \(y_p\)** Guess: \(y_p = Ax + B\) Plug into the original equation: \(y_p' = A\) \(y_p'' = 0\) \[ 2(0) + (A) = 4x \implies A = 4x \] But this can't be right; let's check our guess. The right side is \(4x\), so guess \(y_p = ax + b\): \[ y_p' = a,\quad y_p'' = 0 \] So: \[ 2(0) + (a) = 4x \implies a = 4x \] But \(a\) shouldn't be a function of \(x\). This suggests we need to try a higher degree: \(y_p = ax^2 + bx + c\): \[ y_p' = 2ax + b \] \[ y_p'' = 2a \] Plug into the equation: \[ 2(2a) + (2ax + b) = 4x \implies 4a + 2ax + b = 4x \] Equate coefficients: - \(x\) term: \(2a = 4 \implies a = 2\) - Constant term: \(4a + b = 0 \implies b = -4a = -8\) So, \[ y_p = 2x^2 - 8x \] #### **Step 3: General Solution** \[ y(x) = C_1 + C_2 e^{-x/2} + 2x^2 - 8x \] #### **Step 4: Apply Initial Conditions** \(y(0) = 7\): \[ C_1 + C_2 + 0 - 0 = 7 \implies C_1 + C_2 = 7 \] \(y'(x) = -\frac{1}{2}C_2 e^{-x/2} + 4x - 8\) At \(x=0\): \[ y'(0) = -\frac{1}{2}C_2 + 0 - 8 = -10 \implies -\frac{1}{2}C_2 - 8 = -10 \implies -\frac{1}{2}C_2 = -2 \implies C_2 = 4 \] \[ C_1 = 7 - 4 = 3 \] **Final solution:** \[ \boxed{ y(x) = 3 + 4e^{-x/2} + 2x^2 - 8x } \] --- ### **7. \( y'' - 3y' - 4y = 5e^{2x} \), \( y(0) = 1 \), \( y'(0) = 13 \)** #### **Step 1: Homogeneous Solution** Characteristic equation: \[ r^2 - 3r - 4 = 0 \implies (r - 4)(r + 1) = 0 \implies r = 4,\, -1 \] \[ y_h(x) = C_1 e^{4x} + C_2 e^{-x} \] #### **Step 2: Particular Solution** Guess: \(y_p = Ae^{2x}\) \[ y_p' = 2Ae^{2x} \] \[ y_p'' = 4Ae^{2x} \] Plug in: \[ 4Ae^{2x} - 3(2A e^{2x}) - 4A e^{2x} = 5 e^{2x} \] \[ [4A - 6A - 4A] e^{2x} = 5 e^{2x} \] \[ (-6A) e^{2x} = 5 e^{2x} \implies A = -\frac{5}{6} \] \[ y_p = -\frac{5}{6}e^{2x} \] #### **Step 3: General Solution** \[ y(x) = C_1 e^{4x} + C_2 e^{-x} - \frac{5}{6}e^{2x} \] #### **Step 4: Initial Conditions** \(y(0) = 1\): \[ C_1 + C_2 - \frac{5}{6} = 1 \implies C_1 + C_2 = 1 + \frac{5}{6} = \frac{11}{6} \] \(y'(x) = 4C_1 e^{4x} - C_2 e^{-x} - \frac{10}{6}e^{2x}\) At \(x=0\): \[ y'(0) = 4C_1 - C_2 - \frac{10}{6} = 13 \implies 4C_1 - C_2 = 13 + \frac{10}{6} = \frac{88}{6} \] Solve for \(C_1, C_2\): Add the two equations: 1. \(C_1 + C_2 = \frac{11}{6}\) 2. \(4C_1 - C_2 = \frac{88}{6}\) Add: \[ 4C_1 - C_2 + C_1 + C_2 = \frac{88}{6} + \frac{11}{6} \implies 5C_1 = \frac{99}{6} \implies C_1 = \frac{99}{30} = \frac{33}{10} \] Then, \[ C_2 = \frac{11}{6} - \frac{33}{10} = \frac{55 - 99}{60} = -\frac{44}{60} = -\frac{11}{15} \] **Final solution:** \[ \boxed{ y(x) = \frac{33}{10}e^{4x} - \frac{11}{15}e^{-x} - \frac{5}{6}e^{2x} } \] --- ### **8. \( y'' + 4y = 4\sin(2x) \)** #### **Step 1: Homogeneous Solution** Characteristic equation: \[ r^2 + 4 = 0 \implies r = \pm 2i \] \[ y_h(x) = C_1\cos(2x) + C_2\sin(2x) \] #### **Step 2: Particular Solution** Guess: \(y_p = A\cos(2x) + B\sin(2x)\) \[ y_p' = -2A\sin(2x) + 2B\cos(2x) \] \[ y_p'' = -4A\cos(2x) - 4B\sin(2x) \] Plug in: \[ y_p'' + 4y_p = -4A\cos(2x) - 4B\sin(2x) + 4A\cos(2x) + 4B\sin(2x) = 0 \] So try \(y_p = x(A\cos(2x) + B\sin(2x))\): Try \(y_p = x(C\cos(2x) + D\sin(2x))\): Let’s try \(y_p = x(C\cos(2x) + D\sin(2x))\): Compute derivatives: \(y_p = x[C\cos(2x) + D\sin(2x)]\) \(y_p' = C\cos(2x) + D\sin(2x) + x[-2C\sin(2x) + 2D\cos(2x)]\) \(y_p'' =\) Take derivative term by term: First term: \(C\cos(2x) + D\sin(2x)\) gives: \(-2C\sin(2x) + 2D\cos(2x)\) Second term: \(x[-2C\sin(2x) + 2D\cos(2x)]\) gives: \(-2C\sin(2x) + 2D\cos(2x)\) plus \(x[-4C\cos(2x) - 4D\sin(2x)]\) So add together: \[ y_p'' = -2C\sin(2x) + 2D\cos(2x) -2C\sin(2x) + 2D\cos(2x) + x[-4C\cos(2x) - 4D\sin(2x)] \] Combine terms: \[ y_p'' = -4C\sin(2x) + 4D\cos(2x) + x[-4C\cos(2x) - 4D\sin(2x)] \] Now add \(4y_p\): \[ 4y_p = 4x[C\cos(2x) + D\sin(2x)] \] Total: \[ y_p'' + 4y_p = [-4C\sin(2x) + 4D\cos(2x)] + x[-4C\cos(2x) - 4D\sin(2x) + 4C\cos(2x) + 4D\sin(2x)] \] But notice the \(x\) terms cancel: \[ x[-4C\cos(2x) - 4D\sin(2x) + 4C\cos(2x) + 4D\sin(2x)] = x[0] \] So total: \[ y_p'' + 4y_p = -4C\sin(2x) + 4D\cos(2x) \] Set equal to the right side \(4\sin(2x)\): \[ -4C\sin(2x) + 4D\cos(2x) = 4\sin(2x) \] So, \[ -4C = 4 \implies C = -1 \] \[ 4D = 0 \implies D = 0 \] So, \[ y_p = x(-1)\cos(2x) = -x\cos(2x) \] #### **General Solution** \[ \boxed{ y(x) = C_1\cos(2x) + C_2\sin(2x) - x\cos(2x) } \] --- Let me know if you need **step-by-step explanations** for any individual problem!