Let's solve **Problem 3** step by step: ### **Given:** \[ y'' - 4y' + 4y = -7x + 4x^2, \quad y(0) = 0, \quad y'(0) = 1 \] --- ## **Step 1: Solve the Homogeneous Equation** The homogeneous part is: \[ y'' - 4y' + 4y = 0 \] The characteristic equation: \[ r^2 - 4r + 4 = 0 \implies (r-2)^2 = 0 \implies r = 2 \text{ (double root)} \] So, the general solution to the homogeneous equation is: \[ y_h(x) = (C_1 + C_2 x)e^{2x} \] --- ## **Step 2: Find a Particular Solution** The nonhomogeneous part is \(-7x + 4x^2\). We try a particular solution of the form: \[ y_p(x) = Ax^2 + Bx + C \] Compute derivatives: \[ y_p' = 2Ax + B \] \[ y_p'' = 2A \] Plug into the original equation: \[ y_p'' - 4y_p' + 4y_p = -7x + 4x^2 \] \[ 2A - 4(2Ax + B) + 4(Ax^2 + Bx + C) = -7x + 4x^2 \] \[ 2A - 8Ax - 4B + 4Ax^2 + 4Bx + 4C = -7x + 4x^2 \] Group like terms: - \(x^2\): \(4A x^2\) - \(x\): \(-8A x + 4B x = (-8A + 4B)x\) - Constants: \(2A - 4B + 4C\) So: \[ 4A x^2 + (-8A + 4B)x + (2A - 4B + 4C) = 4x^2 - 7x \] Match coefficients: - \(x^2:\) \(4A = 4 \implies A = 1\) - \(x:\) \(-8A + 4B = -7\) - Plug \(A = 1\): - \(-8(1) + 4B = -7 \implies -8 + 4B = -7 \implies 4B = 1 \implies B = \frac{1}{4}\) - Constants: \(2A - 4B + 4C = 0\) - \(2(1) - 4(\frac{1}{4}) + 4C = 0\) - \(2 - 1 + 4C = 0 \implies 1 + 4C = 0 \implies 4C = -1 \implies C = -\frac{1}{4}\) So, \[ y_p(x) = x^2 + \frac{1}{4}x - \frac{1}{4} \] --- ## **Step 3: General Solution** \[ y(x) = y_h(x) + y_p(x) = (C_1 + C_2 x)e^{2x} + x^2 + \frac{1}{4}x - \frac{1}{4} \] --- ## **Step 4: Apply Initial Conditions** Given: \(y(0) = 0\), \(y'(0) = 1\) ### Compute \(y(0)\): \[ y(0) = (C_1 + C_2 \cdot 0)e^{0} + 0^2 + \frac{1}{4} \cdot 0 - \frac{1}{4} = C_1 - \frac{1}{4} \] Set \(y(0) = 0\): \[ C_1 - \frac{1}{4} = 0 \implies C_1 = \frac{1}{4} \] --- ### Compute \(y'(x)\): \[ y(x) = (C_1 + C_2 x)e^{2x} + x^2 + \frac{1}{4}x - \frac{1}{4} \] Product rule for the first term: \[ \frac{d}{dx}[(C_1 + C_2 x)e^{2x}] = C_2 e^{2x} + (C_1 + C_2 x) \cdot 2e^{2x} = [2C_1 + C_2 + 2C_2 x]e^{2x} \] Derivative of the rest: \(2x + \frac{1}{4}\) So, \[ y'(x) = [2C_1 + C_2 + 2C_2 x]e^{2x} + 2x + \frac{1}{4} \] At \(x = 0\): \[ y'(0) = [2C_1 + C_2 + 0] \cdot 1 + 0 + \frac{1}{4} = 2C_1 + C_2 + \frac{1}{4} \] Set \(y'(0) = 1\) and use \(C_1 = \frac{1}{4}\): \[ 2 \left(\frac{1}{4}\right) + C_2 + \frac{1}{4} = 1 \] \[ \frac{1}{2} + C_2 + \frac{1}{4} = 1 \] \[ C_2 + \frac{3}{4} = 1 \implies C_2 = \frac{1}{4} \] --- ## **Final Answer** \[ \boxed{ y(x) = \left(\frac{1}{4} + \frac{1}{4}x\right)e^{2x} + x^2 + \frac{1}{4}x - \frac{1}{4} } \] --- Let me know if you want solutions for problems 4 or 5!

Let's proceed step-by-step to solve **Problem 4**: --- ### **Given:** \[ y'' + y = e^{x} \] with initial conditions: \[ y(0) = 1, \quad y'(0) = -1 \] --- ## **Step 1: Solve the homogeneous equation** The homogeneous part: \[ y'' + y = 0 \] **Characteristic equation:** \[ r^2 + 1 = 0 \implies r = \pm i \] **Homogeneous solution:** \[ y_h(x) = C_1 \cos x + C_2 \sin x \] --- ## **Step 2: Find a particular solution** The nonhomogeneous term is \(e^{x}\). Since \(e^{x}\) is not a solution to the homogeneous equation, we try a particular solution of the form: \[ y_p(x) = Ae^{x} \] Compute derivatives: \[ y_p' = Ae^{x} \] \[ y_p'' = Ae^{x} \] Plug into the original equation: \[ Ae^{x} + Ae^{x} = e^{x} \] \[ 2A e^{x} = e^{x} \] \[ 2A = 1 \implies A = \frac{1}{2} \] **Particular solution:** \[ y_p(x) = \frac{1}{2} e^{x} \] --- ## **Step 3: General solution** \[ \boxed{ y(x) = C_1 \cos x + C_2 \sin x + \frac{1}{2} e^{x} } \] --- ## **Step 4: Apply initial conditions** ### Compute \(y(0)\): \[ y(0) = C_1 \cos 0 + C_2 \sin 0 + \frac{1}{2} e^{0} = C_1 \cdot 1 + 0 + \frac{1}{2} \cdot 1 = C_1 + \frac{1}{2} \] Given \(y(0) = 1\): \[ C_1 + \frac{1}{2} = 1 \implies C_1 = \frac{1}{2} \] --- ### Compute \(y'(x)\): \[ y'(x) = -C_1 \sin x + C_2 \cos x + \frac{1}{2} e^{x} \] At \(x=0\): \[ y'(0) = -C_1 \cdot 0 + C_2 \cdot 1 + \frac{1}{2} e^{0} = C_2 + \frac{1}{2} \] Given \(y'(0) = -1\): \[ C_2 + \frac{1}{2} = -1 \implies C_2 = -\frac{3}{2} \] --- ## **Final solution:** \[ \boxed{ \boxed{ y(x) = \frac{1}{2} \cos x - \frac{3}{2} \sin x + \frac{1}{2} e^{x} } } \] --- Let me know if you'd like me to proceed with **Problem 5**!