Let's solve each problem step-by-step as requested. --- ## (a) \( y'' + y' + y = -U_{2\pi}(t); \quad y(0) = 2,\, y'(0) = 1 \) ### **Step 1: Solve the Homogeneous Equation** The homogeneous equation is: \[ y'' + y' + y = 0 \] The characteristic equation: \[ r^2 + r + 1 = 0 \] Solve for \( r \): \[ r = \frac{-1 \pm i\sqrt{3}}{2} \] So, the general solution is: \[ y_h(t) = e^{-t/2} \left[ A \cos\left(\frac{\sqrt{3}}{2} t\right) + B \sin\left(\frac{\sqrt{3}}{2} t\right) \right] \] **Explanation:** We solved the homogeneous part using characteristic roots. --- ### **Step 2: Solve for the Particular Solution and Apply Initial Conditions** The nonhomogeneous term is \(-U_{2\pi}(t)\), the unit step at \(2\pi\): - For \(t < 2\pi\): the equation is just the homogeneous one. - For \(t \geq 2\pi\): the equation becomes \(y'' + y' + y = -1\). **For \(t \geq 2\pi\):** Try a particular solution \(y_p = C\). Substitute: \[ 0 + 0 + C = -1 \implies C = -1 \] So, the complete solution for \(t \geq 2\pi\): \[ y(t) = y_h(t) - U_{2\pi}(t) \] Apply initial conditions at \(t = 0\): \[ y(0) = A = 2 \\ y'(0) = -\frac{1}{2}A + \frac{\sqrt{3}}{2}B = 1 \] Using \(A = 2\): \[ -\frac{1}{2} \cdot 2 + \frac{\sqrt{3}}{2}B = 1 \implies -1 + \frac{\sqrt{3}}{2}B = 1 \implies \frac{\sqrt{3}}{2}B = 2 \implies B = \frac{4}{\sqrt{3}} \] **Final Answer:** \[ \boxed{ y(t) = e^{-t/2} \left[2 \cos\left(\frac{\sqrt{3}}{2} t\right) + \frac{4}{\sqrt{3}} \sin\left(\frac{\sqrt{3}}{2} t\right)\right] - U_{2\pi}(t) } \] --- Would you like me to continue with part (b)?

Let's proceed with part (b) step-by-step. --- ## (b) \( y'' + 4y = \delta_0(t); \quad y(0) = 0, \quad y'(0) = 0 \) ### **Step 1: Find the homogeneous solution** The homogeneous equation is: \[ y'' + 4y = 0 \] Characteristic equation: \[ r^2 + 4 = 0 \implies r = \pm 2i \] Solution: \[ y_h(t) = C_1 \cos 2t + C_2 \sin 2t \] **Explanation:** This is a standard second-order linear differential equation with constant coefficients, and the roots are purely imaginary, leading to sinusoidal solutions. --- ### **Step 2: Find the particular solution considering the impulse at \( t=0 \)** Since the forcing term is \(\delta_0(t)\), an impulse at \(t=0\), the solution involves an impulse response. The general approach: - The impulse at \(t=0\) causes a discontinuity in the derivative \(y'\) but not in \(y\). - Integrate the differential equation across \(t=0\) to find the jump in \(y'\). **Integrate both sides over \([- \varepsilon, \varepsilon]\):** \[ \int_{-\varepsilon}^{\varepsilon} y''(t) dt + 4 \int_{-\varepsilon}^{\varepsilon} y(t) dt = \int_{-\varepsilon}^{\varepsilon} \delta(t) dt \] \[ [y'(\varepsilon) - y'(-\varepsilon)] + 4 \cdots = 1 \] Since \(y(t)\) is finite and continuous at \(t=0\), as \(\varepsilon \to 0\): \[ [y'(0^+) - y'(0^-)] = 1 \] and \(y(0^+) = y(0^-) = 0\). Given initial conditions: \[ y(0) = 0,\quad y'(0^-) = 0 \] and the jump: \[ y'(0^+) = y'(0^-) + 1 = 1 \] **Solution:** - For \(t > 0\), the solution satisfies: \[ y(t) = A \cos 2t + B \sin 2t \] with initial conditions: \[ y(0) = 0 \implies A = 0 \] \[ y'(t) = -2A \sin 2t + 2B \cos 2t \] At \(t=0\): \[ y'(0^+) = 2B = 1 \implies B = \frac{1}{2} \] **Final solution:** \[ \boxed{ y(t) = \frac{1}{2} \sin 2t \quad \text{for } t \geq 0 } \] --- **Summary:** - The homogeneous solution describes oscillations at frequency 2. - The impulse at \(t=0\) causes an initial velocity \( y'(0^+) = 1 \). - The resulting solution for \(t \geq 0\): \[ \boxed{ y(t) = \frac{1}{2} \sin 2t } \] --- Would you like me to clarify further or proceed with any additional parts?

Let's analyze and solve the initial value problems step-by-step. --- ### (a) \( y'' + y' + y = -U_{2\pi}(t); \quad y(0)=2,\ y'(0)=1 \) **Step 1: Solve the homogeneous equation** The homogeneous differential equation: \[ y'' + y' + y = 0 \] Characteristic equation: \[ r^2 + r + 1 = 0 \] Solutions: \[ r = \frac{-1 \pm i\sqrt{3}}{2} \] which are complex conjugates. The general homogeneous solution: \[ y_h(t) = e^{-t/2}\left[ A \cos\left(\frac{\sqrt{3}}{2} t\right) + B \sin\left(\frac{\sqrt{3}}{2} t\right) \right] \] **Explanation:** We found the roots of the characteristic equation and expressed the solution in exponential form with sinusoidal components. --- **Step 2: Handle the forcing term and apply initial conditions** The forcing term is the step function \(-U_{2\pi}(t)\), which is zero for \(t<2\pi\) and \(-1\) for \(t \ge 2\pi\). - For \(t < 2\pi\): the equation reduces to the homogeneous one, with initial conditions as given. - For \(t \ge 2\pi\): the equation becomes: \[ y'' + y' + y = -1 \] **Initial conditions at \(t=0\):** \[ y(0) = 2 \implies \text{from } y_h(0):\ A = 2 \] \[ y'(t) = -\frac{1}{2} e^{-t/2}\left[ A \cos\left(\frac{\sqrt{3}}{2} t\right) + B \sin\left(\frac{\sqrt{3}}{2} t\right) \right] + e^{-t/2}\left[ -A \frac{\sqrt{3}}{2} \sin\left(\frac{\sqrt{3}}{2} t\right) + B \frac{\sqrt{3}}{2} \cos\left(\frac{\sqrt{3}}{2} t\right) \right] \] At \(t=0\): \[ y'(0) = -\frac{1}{2} \times 1 \times A + 1 \times B \frac{\sqrt{3}}{2} = 1 \] Substituting \(A=2\): \[ -\frac{1}{2} \times 2 + \frac{\sqrt{3}}{2} B = 1 \implies -1 + \frac{\sqrt{3}}{2} B = 1 \] \[ \Rightarrow \frac{\sqrt{3}}{2} B = 2 \implies B = \frac{4}{\sqrt{3}} \] **Particular solution for \(t \ge 2\pi\):** Since the RHS is \(-1\), try a constant particular solution \(y_p = C\): \[ 0 + 0 + C = -1 \Rightarrow C = -1 \] Thus, for \(t \ge 2\pi\): \[ y(t) = y_h(t) - 1 \] **Final solution:** \[ \boxed{ y(t) = e^{-t/2}\left[ 2 \cos\left(\frac{\sqrt{3}}{2} t\right) + \frac{4}{\sqrt{3}} \sin\left(\frac{\sqrt{3}}{2} t\right) \right] - U_{2\pi}(t) } \] This combines the homogeneous solution with the step response, satisfying initial conditions at \(t=0\). --- ### (b) \( y'' + 4 y = \delta(t); \quad y(0) = 0,\ y'(0) = 0 \) **Step 1: Homogeneous solution** \[ y'' + 4 y = 0 \] Characteristic roots: \[ r = \pm 2 i \] Solution: \[ y_h(t) = C_1 \cos 2t + C_2 \sin 2t \] **Step 2: Effect of the impulse at \(t=0\)** The delta function causes an instantaneous change in the derivative \(y'\), while \(y\) remains continuous. Integrate the equation over an infinitesimal interval around \(t=0\): \[ \int_{-\varepsilon}^{\varepsilon} y'' + 4 y \, dt = \int_{-\varepsilon}^{\varepsilon} \delta(t) dt \] \[ [y'(\varepsilon) - y'(-\varepsilon)] + 4 \int_{-\varepsilon}^{\varepsilon} y(t) dt = 1 \] As \(\varepsilon \to 0\), \(y(t)\) is finite and continuous, so: \[ [y'(0^+) - y'(0^-)] = 1 \] with initial \(y(0) = 0\), \(y'(0^-) = 0\). Therefore: \[ y(0) = 0 \Rightarrow A = 0 \] \[ y'(t) = -2 C_1 \sin 2t + 2 C_2 \cos 2t \] At \(t=0\): \[ y'(0^+) = 2 C_2 = 1 \Rightarrow C_2 = \frac{1}{2} \] Since initial \(y(0) = 0\), \(A=0\), and initial derivative jumps to 1. The solution for \(t > 0\): \[ y(t) = \frac{1}{2} \sin 2t \] --- ### **Final answer:** \[ \boxed{ \text{For } t \ge 0,\quad y(t) = \frac{1}{2} \sin 2t } \] This solution captures the effect of an instantaneous impulse at \(t=0\), producing an initial velocity of 1, with \(y(0)=0\). --- If you need further clarification or additional parts, feel free to ask!