Let's solve each initial value problem step by step, explaining at each stage and providing comments on the initial conditions and realism. --- ## (a) ### Equation: \( y''(t) + 3y'(t) - 4y(t) = 2e^{2t} + 3 \), with \( y(0) = 2 \), \( y'(0) = 1 \). #### Step 1: Solve the homogeneous equation and find the general solution The homogeneous equation is: \( y'' + 3y' - 4y = 0 \) **Characteristic equation:** \( r^2 + 3r - 4 = 0 \) \( (r + 4)(r - 1) = 0 \) So, \( r = -4, 1 \) **General solution to the homogeneous equation:** \( y_h(t) = C_1 e^{-4t} + C_2 e^{t} \) #### Step 1 Explanation: We find the solution to the system if there were no external force (the right-hand side is zero). The characteristic equation gives us two real roots, meaning the system responds with exponential decay and exponential growth (which is a sign that the system is not physically stable). --- #### Step 2: Find a particular solution to the nonhomogeneous equation Let’s try a particular solution of the form: \( y_p(t) = Ae^{2t} + B \) (since the right-hand side is \( 2e^{2t} + 3 \)) Calculate derivatives: - \( y_p' = 2A e^{2t} \) - \( y_p'' = 4A e^{2t} \) Plug into the equation: \[ y_p'' + 3y_p' - 4y_p = [4A e^{2t} + 3(2A e^{2t}) - 4(A e^{2t} + B)] = 4A e^{2t} + 6A e^{2t} - 4A e^{2t} - 4B = 6A e^{2t} - 4B \] Set equal to the forcing term: \( 6A e^{2t} - 4B = 2e^{2t} + 3 \) So, - \( 6A = 2 \implies A = \frac{1}{3} \) - \( -4B = 3 \implies B = -\frac{3}{4} \) So, \( y_p(t) = \frac{1}{3}e^{2t} - \frac{3}{4} \) **General solution:** \[ y(t) = C_1 e^{-4t} + C_2 e^{t} + \frac{1}{3}e^{2t} - \frac{3}{4} \] Now, use initial conditions: - \( y(0) = C_1 + C_2 + \frac{1}{3} - \frac{3}{4} = 2 \) - \( y'(t) = -4C_1 e^{-4t} + C_2 e^{t} + \frac{2}{3}e^{2t} \) - \( y'(0) = -4C_1 + C_2 + \frac{2}{3} = 1 \) Set up equations: 1. \( C_1 + C_2 + \frac{1}{3} - \frac{3}{4} = 2 \implies C_1 + C_2 = 2 + \frac{3}{4} - \frac{1}{3} = 2 + \frac{9-4}{12} = 2 + \frac{5}{12} = \frac{29}{12} \) 2. \( -4C_1 + C_2 + \frac{2}{3} = 1 \implies -4C_1 + C_2 = 1 - \frac{2}{3} = \frac{1}{3} \) Subtract equation 1 from equation 2: Let’s solve: From 1: \( C_2 = \frac{29}{12} - C_1 \) Substitute into 2: \[ -4C_1 + \left( \frac{29}{12} - C_1 \right) = \frac{1}{3} \\ -4C_1 - C_1 + \frac{29}{12} = \frac{1}{3} \\ -5C_1 = \frac{1}{3} - \frac{29}{12} \\ = \frac{4}{12} - \frac{29}{12} = -\frac{25}{12} \\ C_1 = \frac{5}{12 \times 5} = \frac{5}{60} = \frac{1}{12} \] Wait, but \( -5C_1 = -\frac{25}{12} \implies C_1 = \frac{25}{60} = \frac{5}{12} \). So \( C_1 = \frac{5}{12} \), \( C_2 = \frac{29}{12} - \frac{5}{12} = \frac{24}{12} = 2 \). So, \[ \boxed{ y(t) = \frac{5}{12}e^{-4t} + 2e^{t} + \frac{1}{3}e^{2t} - \frac{3}{4} } \] #### Step 2 Explanation: We found a particular solution using the method of undetermined coefficients, then applied the initial conditions to find the specific constants for this problem. --- **Initial conditions meaning:** - \( y(0) = 2 \): The initial displacement is 2 units upward from equilibrium. - \( y'(0) = 1 \): The initial velocity is 1 unit per time upward. **Comment on realism:** - The presence of an exponentially growing mode (\( e^{t} \), \( e^{2t} \)) means the system is unstable—real spring-mass systems should have all negative real parts in the characteristic equation (all modes decay). The negative coefficient on the \( y \) term (spring constant) and positive on the damping term make this physically unrealistic (a spring shouldn't have negative stiffness). **Final Answer for (a):** \[ \boxed{ y(t) = \frac{5}{12}e^{-4t} + 2e^{t} + \frac{1}{3}e^{2t} - \frac{3}{4} } \] --- ## (b) ### Equation: \( y''(t) - 5y'(t) - 6y(t) = 3t - t - 2 \), with \( y(0) = 0 \), \( y'(0) = 1 \). #### Step 1: Simplify and solve the homogeneous equation Simplify the right side: \( 3t - t - 2 = 2t - 2 \) So, \( y'' - 5y' - 6y = 2t - 2 \) Homogeneous equation: \( y'' - 5y' - 6y = 0 \) Characteristic equation: \( r^2 - 5r - 6 = 0 \) \( (r - 6)(r + 1) = 0 \) \( r = 6, -1 \) General homogeneous solution: \( y_h = C_1 e^{6t} + C_2 e^{-t} \) #### Step 1 Explanation: We find the natural response modes, which are one growing exponentially and one decaying exponentially; again, exponential growth is not physically realistic for a damped spring. --- #### Step 2: Find a particular solution Try \( y_p = At + B \). - \( y_p' = A \) - \( y_p'' = 0 \) Plug in: \[ 0 - 5A - 6(At + B) = 2t - 2 \\ -5A - 6At - 6B = 2t - 2 \] Group by powers of \( t \): - Coefficient of \( t \): \( -6A t = 2t \implies A = -\frac{1}{3} \) - Constant: \( -5A - 6B = -2 \) Substitute \( A = -\frac{1}{3} \): \( -5(-\frac{1}{3}) - 6B = -2 \implies \frac{5}{3} - 6B = -2 \implies -6B = -2 - \frac{5}{3} = -\frac{11}{3} \implies B = \frac{11}{18} \) So, \( y_p = -\frac{1}{3}t + \frac{11}{18} \) General solution: \[ y(t) = C_1 e^{6t} + C_2 e^{-t} - \frac{1}{3}t + \frac{11}{18} \] Apply initial conditions: - \( y(0) = C_1 + C_2 + \frac{11}{18} = 0 \implies C_1 + C_2 = -\frac{11}{18} \) - \( y'(t) = 6C_1 e^{6t} - C_2 e^{-t} - \frac{1}{3} \) - \( y'(0) = 6C_1 - C_2 - \frac{1}{3} = 1 \) From above: 1. \( C_1 + C_2 = -\frac{11}{18} \) 2. \( 6C_1 - C_2 = 1 + \frac{1}{3} = \frac{4}{3} \) Add equations: \( 6C_1 - C_2 = \frac{4}{3} \) \( C_1 + C_2 = -\frac{11}{18} \) Let’s solve: Add the two: First, from the second equation: \( C_2 = -\frac{11}{18} - C_1 \) Substitute into the first: \[ 6C_1 - (-\frac{11}{18} - C_1) = \frac{4}{3} \\ 6C_1 + C_1 + \frac{11}{18} = \frac{4}{3} \\ 7C_1 = \frac{4}{3} - \frac{11}{18} = \frac{24-11}{18} = \frac{13}{18} \\ C_1 = \frac{13}{18 \times 7} = \frac{13}{126} \] Then, \( C_2 = -\frac{11}{18} - \frac{13}{126} = -\frac{77}{126} - \frac{13}{126} = -\frac{90}{126} = -\frac{15}{21} = -\frac{5}{7} \) So, \[ \boxed{ y(t) = \frac{13}{126}e^{6t} - \frac{5}{7}e^{-t} - \frac{1}{3}t + \frac{11}{18} } \] #### Step 2 Explanation: We found a particular solution and solved for the constants using the initial conditions. --- **Initial conditions meaning:** - \( y(0) = 0 \): The mass starts at equilibrium. - \( y'(0) = 1 \): The mass is initially moving upward at velocity 1. **Comment on realism:** - The positive root in the characteristic equation (\( r = 6 \)) means the solution can grow without bound, not physically realistic for a damped spring-mass system, which should have all negative real parts in the characteristic equation. **Final Answer for (b):** \[ \boxed{ y(t) = \frac{13}{126}e^{6t} - \frac{5}{7}e^{-t} - \frac{1}{3}t + \frac{11}{18} } \] --- ## (c) ### Equation: \( y''(t) - 5y'(t) + 6y(t) = 3t - t \), \( y(0) = 2 \), \( y'(0) = -1 \) #### Step 1: Simplify and solve the homogeneous equation Right side: \( 3t - t = 2t \) Homogeneous equation: \( y'' - 5y' + 6y = 0 \) Characteristic equation: \( r^2 - 5r + 6 = 0 \implies (r-2)(r-3) = 0 \implies r = 2, 3 \) General solution: \( y_h = C_1 e^{2t} + C_2 e^{3t} \) #### Step 1 Explanation: Both roots are positive, so both modes grow exponentially; this is not realistic for a stable spring-mass system. --- #### Step 2: Find a particular solution Try \( y_p = At + B \) - \( y_p' = A \) - \( y_p'' = 0 \) Plug in: \( 0 - 5A + 6(At + B) = 2t \) \( -5A + 6A t + 6B = 2t \) Match coefficients: - \( 6A t = 2t \implies A = \frac{1}{3} \) - \( -5A + 6

Let's solve each initial value problem step-by-step, providing explanations after each step and concluding with the final answer. --- ### **Part (a)** **Given:** \[ y'' + 3y' - 4y = 2e^{2t} + 3,\quad y(0)=2,\quad y'(0)=1 \] --- ### **Step 1:** Find the homogeneous solution **Characteristic equation:** \[ r^2 + 3r - 4 = 0 \] Factor: \[ (r + 4)(r - 1) = 0 \implies r = -4, \quad r=1 \] **Homogeneous solution:** \[ y_h(t) = C_1 e^{-4t} + C_2 e^{t} \] *Explanation:* The homogeneous solution describes how the system would behave without external forcing. The roots indicate exponential decay (\(e^{-4t}\)) and growth (\(e^{t}\)), with the system naturally tending to decay for the \(e^{-4t}\) mode, but the \(e^{t}\) mode indicates potential instability. --- ### **Step 2:** Find a particular solution Since the forcing includes \(2e^{2t}\) (exponential) and a constant, try: \[ y_p(t) = A e^{2t} + B \] Calculate derivatives: \[ y_p' = 2A e^{2t}, \quad y_p'' = 4A e^{2t} \] Substitute into the original equation: \[ 4A e^{2t} + 3(2A e^{2t}) - 4(A e^{2t} + B) = 2 e^{2t} + 3 \] Simplify: \[ 4A e^{2t} + 6A e^{2t} - 4A e^{2t} - 4B = 2 e^{2t} + 3 \] Combine like terms: \[ (4A + 6A - 4A) e^{2t} - 4B = 2 e^{2t} + 3 \implies 6A e^{2t} - 4B = 2 e^{2t} + 3 \] Matching coefficients: - For \(e^{2t}\): \[ 6A = 2 \implies A = \frac{1}{3} \] - For constants: \[ -4B = 3 \implies B = -\frac{3}{4} \] **Particular solution:** \[ y_p(t) = \frac{1}{3} e^{2t} - \frac{3}{4} \] **General solution:** \[ \boxed{ y(t) = C_1 e^{-4t} + C_2 e^{t} + \frac{1}{3} e^{2t} - \frac{3}{4} } \] --- ### **Step 3:** Apply initial conditions to find \(C_1, C_2\) Calculate \( y(0) \): \[ C_1 + C_2 + \frac{1}{3} - \frac{3}{4} = 2 \] Simplify: \[ C_1 + C_2 = 2 - \frac{1}{3} + \frac{3}{4} \] Express all with denominator 12: \[ 2 = \frac{24}{12}, \quad \frac{1}{3} = \frac{4}{12}, \quad \frac{3}{4} = \frac{9}{12} \] So: \[ C_1 + C_2 = \frac{24}{12} - \frac{4}{12} + \frac{9}{12} = \frac{29}{12} \] Calculate \( y'(t) \): \[ y' = -4 C_1 e^{-4t} + C_2 e^{t} + \frac{2}{3} e^{2t} \] At \( t=0 \): \[ -4 C_1 + C_2 + \frac{2}{3} = 1 \] Express \(\frac{2}{3}\) as \(\frac{8}{12}\), and rewrite the first equation: \[ C_2 = \frac{29}{12} - C_1 \] Substitute into the second: \[ -4 C_1 + \left( \frac{29}{12} - C_1 \right) + \frac{8}{12} = 1 \] Simplify: \[ -4 C_1 + \frac{29}{12} - C_1 + \frac{8}{12} = 1 \] \[ (-4 C_1 - C_1) + \left( \frac{29}{12} + \frac{8}{12} \right) = 1 \] \[ -5 C_1 + \frac{37}{12} = 1 \] Express 1 as \(\frac{12}{12}\): \[ -5 C_1 = \frac{12}{12} - \frac{37}{12} = -\frac{25}{12} \] \[ C_1 = \frac{25}{12 \times 5} = \frac{25}{60} = \frac{5}{12} \] Then: \[ C_2 = \frac{29}{12} - \frac{5}{12} = \frac{24}{12} = 2 \] --- ### **Final answer for (a):** \[ \boxed{ \boxed{ y(t) = \frac{5}{12} e^{-4t} + 2 e^{t} + \frac{1}{3} e^{2t} - \frac{3}{4} } } \] --- ## **Part (a) summary:** - **Initial conditions:** \( y(0)=2 \) indicates an initial displacement of 2 units above equilibrium; \( y'(0)=1 \) indicates initial upward velocity of 1 unit/sec. - **Comments on the solution:** The presence of \( e^{t} \) and \( e^{2t} \) terms suggests the system exhibits exponential growth, which is physically unrealistic for a typical spring-mass system with damping. Real systems should have all modes decay or oscillate without unbounded growth. --- **Proceeding similarly for parts (b) and (c):** (If you want, I can continue with parts (b) and (c) following the same two-step process.)

Certainly! Let's proceed with parts (b) and (c) following the same two-step approach, with explanations after each step and a final summary. --- ## **Part (b)** **Given:** \[ y'' - 5 y' - 6 y = 3t - t - 2 = 2t - 2,\quad y(0)=0,\quad y'(0)=1 \] --- ### **Step 1:** Find the homogeneous solution **Characteristic equation:** \[ r^2 - 5r - 6 = 0 \] Factor: \[ (r - 6)(r + 1) = 0 \implies r=6,\; r=-1 \] **Homogeneous solution:** \[ y_h(t) = C_1 e^{6t} + C_2 e^{-t} \] *Explanation:* Both roots are real, with one positive (growth) and one negative (decay). The positive exponential indicates an unstable growth mode, which is not typical in a physically damped spring-mass system. --- ### **Step 2:** Find a particular solution Since the forcing term is \( 2t - 2 \), a polynomial of degree 1, try: \[ y_p(t) = A t + B \] Derivatives: \[ y_p' = A,\quad y_p''=0 \] Substitute into the differential equation: \[ 0 - 5A - 6 (A t + B) = 2t - 2 \] Simplify: \[ -5A - 6A t - 6B = 2 t - 2 \] Matching coefficients: - For \( t \): \[ -6A = 2 \implies A = -\frac{1}{3} \] - For constants: \[ -5A - 6B = -2 \] Substitute \( A = -\frac{1}{3} \): \[ -5 \left(-\frac{1}{3}\right) - 6B = -2 \implies \frac{5}{3} - 6B = -2 \] Solve for \( B \): \[ -6B = -2 - \frac{5}{3} = -\frac{6}{3} - \frac{5}{3} = -\frac{11}{3} \] \[ B = \frac{11}{18} \] **Particular solution:** \[ y_p(t) = -\frac{1}{3} t + \frac{11}{18} \] **General solution:** \[ \boxed{ y(t) = C_1 e^{6t} + C_2 e^{-t} - \frac{1}{3} t + \frac{11}{18} } \] --- ### **Step 3:** Use initial conditions to find \( C_1, C_2 \) Calculate \( y(0) \): \[ C_1 + C_2 + \frac{11}{18} = 0 \implies C_1 + C_2 = - \frac{11}{18} \] Calculate \( y'(t) \): \[ y' = 6 C_1 e^{6t} - C_2 e^{-t} - \frac{1}{3} \] At \( t=0 \): \[ 6 C_1 - C_2 - \frac{1}{3} = 1 \] Express \( C_2 \): \[ C_2 = - \frac{11}{18} - C_1 \] Substitute into the second: \[ 6 C_1 - \left( - \frac{11}{18} - C_1 \right) - \frac{1}{3} = 1 \] Simplify: \[ 6 C_1 + \frac{11}{18} + C_1 - \frac{1}{3} = 1 \] \[ (6 C_1 + C_1) + \left( \frac{11}{18} - \frac{6}{18} \right) = 1 \] \[ 7 C_1 + \frac{5}{18} = 1 \] Express 1 as \( \frac{18}{18} \): \[ 7 C_1 = \frac{18}{18} - \frac{5}{18} = \frac{13}{18} \] \[ C_1 = \frac{13}{18 \times 7} = \frac{13}{126} \] And \[ C_2 = - \frac{11}{18} - \frac{13}{126} \] Express \( - \frac{11}{18} = - \frac{77}{126} \): \[ C_2 = - \frac{77}{126} - \frac{13}{126} = - \frac{90}{126} = - \frac{15}{21} = - \frac{5}{7} \] --- ### **Final answer for (b):** \[ \boxed{ \boxed{ y(t) = \frac{13}{126} e^{6t} - \frac{5}{7} e^{-t} - \frac{1}{3} t + \frac{11}{18} } } \] --- ### **Summary for (b):** - **Initial conditions:** \( y(0)=0 \): starting at equilibrium, and \( y'(0)=1 \): initial upward velocity. - **Comments on the solution:** The mode \( e^{6t} \) indicates exponential growth, which suggests an unstable system. Such growth is physically unrealistic for a damped spring-mass, but mathematically it is correct given the roots. --- ## **Part (c)** **Given:** \[ y'' - 5 y' + 6 y = 2t,\quad y(0)=2,\quad y'(0)=-1 \] --- ### **Step 1:** Homogeneous solution Characteristic equation: \[ r^2 - 5 r + 6 = 0 \implies (r-2)(r-3)=0 \] Roots: \[ r=2,\quad r=3 \] **Homogeneous solution:** \[ y_h(t) = C_1 e^{2t} + C_2 e^{3t} \] *Explanation:* Both roots positive, indicating solutions grow exponentially over time, which is typical of an unstable mode if uncontrolled. --- ### **Step 2:** Find a particular solution Since the forcing \( 2t \) is linear, try: \[ y_p(t) = A t + B \] Derivatives: \[ y_p' = A,\quad y_p''=0 \] Substitute into the differential equation: \[ 0 - 5A + 6(A t + B) = 2 t \] Simplify: \[ -5A + 6A t + 6 B = 2 t \] Matching coefficients: - For \( t \): \( 6A = 2 \implies A = \frac{1}{3} \) - For constants: \( -5A + 6 B = 0 \) Substitute \( A = \frac{1}{3} \): \[ -\frac{5}{3} + 6 B = 0 \implies 6 B = \frac{5}{3} \implies B = \frac{5}{18} \] **Particular solution:** \[ y_p(t) = \frac{1}{3} t + \frac{5}{18} \] **General solution:** \[ \boxed{ y(t) = C_1 e^{2t} + C_2 e^{3t} + \frac{1}{3} t + \frac{5}{18} } \] --- ### **Step 3:** Apply initial conditions Calculate \( y(0) \): \[ C_1 + C_2 + \frac{5}{18} = 2 \] \[ C_1 + C_2 = 2 - \frac{5}{18} = \frac{36}{18} - \frac{5}{18} = \frac{31}{18} \] Calculate \( y'(t) \): \[ y' = 2 C_1 e^{2t} + 3 C_2 e^{3t} + \frac{1}{3} \] At \( t=0 \): \[ 2 C_1 + 3 C_2 + \frac{1}{3} = -1 \] Express \( C_2 \): \[ C_2 = \frac{31}{18} - C_1 \] Substitute: \[ 2 C_1 + 3 \left( \frac{31}{18} - C_1 \right) + \frac{1}{3} = -1 \] Simplify: \[ 2 C_1 + \frac{93}{18} - 3 C_1 + \frac{1}{3} = -1 \] \[ (2 C_1 - 3 C_1) + \left( \frac{93}{18} + \frac{6}{18} \right) = -1 \] \[ - C_1 + \frac{99}{18} = -1 \] Express \(-1\) as \(- \frac{18}{18}\): \[ - C_1 = - \frac{18}{18} - \frac{99}{18} = - \frac{117}{18} \] \[ C_1 = \frac{117}{18} = \frac{13}{2} \] Then, \[ C_2 = \frac{31}{18} - \frac{13}{2} = \frac{31}{18} - \frac{117}{18} = - \frac{86}{18} = - \frac{43}{9} \] --- ### **Final answer for (c):** \[ \boxed{ \boxed{ y(t) = \frac{13}{2} e^{2t} - \frac{43}{9} e^{3t} + \frac{1}{3} t + \frac{5}{18} } } \] --- ## **Summary for (c):** - **Initial conditions:** \( y(0)=2 \) (initial displacement), \( y'(0)=-1 \) (initial downward velocity). - **Comments on the solution:** The exponential terms grow rapidly, indicating an unstable system—physically unrealistic for a stable spring-mass but correct mathematically. --- ## **Overall Conclusion:** - The solutions involve exponential functions reflecting natural modes of the system. - The presence of positive exponentials suggests instability, which is often non-physical for real spring-mass systems with damping. - Initial conditions specify starting displacement and velocity, directly influencing the particular solution constants. - These mathematical solutions help understand the system behavior but must be interpreted carefully regarding physical realism.