Let's solve each initial value problem **step by step**: --- ### (a) \( y'' + y = 0, \quad y(\pi/3) = 2, \quad y'(\pi/3) = -4 \) #### **Step 1: Solve the homogeneous equation** Characteristic equation: \[ r^2 + 1 = 0 \implies r = \pm i \] General solution: \[ y(t) = A \cos t + B \sin t \] #### **Step 2: Find derivatives** \[ y'(t) = -A \sin t + B \cos t \] #### **Step 3: Apply initial conditions** At \( t = \pi/3 \): \[ y(\pi/3) = A \cos(\pi/3) + B \sin(\pi/3) = 2 \] \[ y'(\pi/3) = -A \sin(\pi/3) + B \cos(\pi/3) = -4 \] Recall: \[ \cos(\pi/3) = \frac{1}{2}, \quad \sin(\pi/3) = \frac{\sqrt{3}}{2} \] So: \[ A \cdot \frac{1}{2} + B \cdot \frac{\sqrt{3}}{2} = 2 \quad \text{(1)} \] \[ -A \cdot \frac{\sqrt{3}}{2} + B \cdot \frac{1}{2} = -4 \quad \text{(2)} \] #### **Step 4: Solve for A and B** Multiply (1) by 2: \[ A + B\sqrt{3} = 4 \] Multiply (2) by 2: \[ -A\sqrt{3} + B = -8 \] Now solve the system: From (1): \[ A = 4 - B\sqrt{3} \] Plug into (2): \[ -(4 - B\sqrt{3})\sqrt{3} + B = -8 \\ -4\sqrt{3} + 3B + B = -8 \\ -4\sqrt{3} + 4B = -8 \\ 4B = -8 + 4\sqrt{3} \\ B = -2 + \sqrt{3} \] Now, \(A = 4 - (-2 + \sqrt{3})\sqrt{3} = 4 + 2\sqrt{3} - 3 = 1 + 2\sqrt{3}\) #### **Final Answer (a):** \[ \boxed{ y(t) = (1 + 2\sqrt{3}) \cos t + (-2 + \sqrt{3}) \sin t } \] --- ### (b) \( y'' - 2y' + 5y = 0, \quad y(\pi/2) = 0, \quad y'(\pi/2) = 2 \) #### **Step 1: Characteristic equation** \[ r^2 - 2r + 5 = 0 \implies r = 1 \pm 2i \] #### **Step 2: General solution** \[ y(t) = e^{t} (A \cos 2t + B \sin 2t) \] #### **Step 3: Find derivatives** \[ y'(t) = e^{t}[A \cos 2t + B \sin 2t]' + e^{t}[A \cos 2t + B \sin 2t] \] Using product rule: \[ y'(t) = e^{t}[ -2A \sin 2t + 2B \cos 2t ] + e^{t}[A \cos 2t + B \sin 2t] \] \[ = e^{t}[A \cos 2t + B \sin 2t - 2A \sin 2t + 2B \cos 2t] \] \[ = e^{t}[(A + 2B)\cos 2t + (B - 2A)\sin 2t] \] #### **Step 4: Apply initial conditions \( t = \pi/2 \)** \[ y(\pi/2) = e^{\pi/2}[A \cos \pi + B \sin \pi] = 0 \implies e^{\pi/2}[A \cdot (-1) + B \cdot 0] = 0 \implies -A e^{\pi/2} = 0 \implies A = 0 \] Now, \( y'(t) = e^{t}[2B\cos 2t - 2A\sin 2t + B\sin 2t + A\cos 2t] \), with \(A = 0\): \[ y'(t) = e^{t}[2B \cos 2t + B \sin 2t] \] At \( t = \pi/2 \): \[ y'(\pi/2) = e^{\pi/2}[2B \cos \pi + B \sin \pi] = e^{\pi/2}[2B \cdot (-1) + B \cdot 0] = -2B e^{\pi/2} \] Set equal to 2: \[ -2B e^{\pi/2} = 2 \implies B = -\frac{1}{e^{\pi/2}} \] #### **Final Answer (b):** \[ \boxed{ y(t) = -\frac{1}{e^{\pi/2}} e^{t} \sin 2t } \] --- ### (c) \( y''' + 12y'' + 36y' = 0, \quad y(0) = 0, \quad y'(0) = 1, \quad y''(0) = 6 \) #### **Step 1: Characteristic equation** \[ r^3 + 12r^2 + 36r = 0 \implies r(r^2 + 12r + 36) = 0 \implies r = 0, r = -6, r = -6 \] #### **Step 2: General solution** \[ y(t) = A + B e^{-6t} + C t e^{-6t} \] #### **Step 3: Find derivatives** First derivative: \[ y'(t) = -6B e^{-6t} + C e^{-6t} - 6C t e^{-6t} \] \[ = (-6B + C) e^{-6t} - 6C t e^{-6t} \] Second derivative: \[ y''(t) = [(-6B + C)(-6) e^{-6t}] + [-6C e^{-6t} - 6C(-6t) e^{-6t}] \] \[ = [36B - 6C - 6C] e^{-6t} + 36C t e^{-6t} \] \[ = [36B - 12C] e^{-6t} + 36C t e^{-6t} \] #### **Step 4: Apply initial conditions at \( t = 0 \)** \( y(0) = A + B = 0 \implies A = -B \) \( y'(0) = (-6B + C) = 1 \) \( y''(0) = 36B - 12C = 6 \) Now solve the system: From above: 1. \( A + B = 0 \implies A = -B \) 2. \( -6B + C = 1 \) 3. \( 36B - 12C = 6 \) From (2): \( C = 1 + 6B \) Plug into (3): \[ 36B - 12(1 + 6B) = 6 \\ 36B - 12 - 72B = 6 \\ -36B = 18 \implies B = -\frac{1}{2} \] Then \( A = -B = \frac{1}{2} \) From (2): \( C = 1 + 6(-\frac{1}{2}) = 1 - 3 = -2 \) #### **Final Answer (c):** \[ \boxed{ y(t) = \frac{1}{2} - \frac{1}{2} e^{-6t} - 2t e^{-6t} } \] --- ### **Behavior for Increasing \( t \) and Sketch** #### (a) **Oscillatory** (sine/cosine terms): Bounded, periodic solution. #### (b) **Oscillatory with Exponential Growth**: Amplitude grows as \( e^t \), oscillatory due to sine. #### (c) **Converges to Constant**: As \( t \to \infty \), \( e^{-6t} \to 0 \), so \( y(t) \to \frac{1}{2} \). #### **Graph Summary:** - (a): Pure sine/cosine, oscillates forever. - (b): Oscillates but grows without bound as \( t \to \infty \). - (c): Quickly approaches \( \frac{1}{2} \) as \( t \to \infty \). If you need plots, let me know!

Let's carefully solve each initial value problem (IVP) step-by-step, including detailed calculations, and then summarize the final solutions. --- ## (a) \( y'' + y = 0 \), with \( y(\pi/3) = 2 \), \( y'(\pi/3) = -4 \) ### Step 1: Solve the homogeneous differential equation Characteristic equation: \[ r^2 + 1 = 0 \implies r = \pm i \] General solution: \[ \boxed{ y(t) = A \cos t + B \sin t } \] ### Step 2: Find derivatives \[ y'(t) = -A \sin t + B \cos t \] ### Step 3: Apply initial conditions at \( t = \pi/3 \) Given: \[ y(\pi/3) = 2 \] \[ y'(\pi/3) = -4 \] Substitute \( t = \pi/3 \): \[ A \cos (\pi/3) + B \sin (\pi/3) = 2 \] \[ -A \sin (\pi/3) + B \cos (\pi/3) = -4 \] Recall: \[ \cos (\pi/3) = \frac{1}{2}, \quad \sin (\pi/3) = \frac{\sqrt{3}}{2} \] Plugging in: \[ A \cdot \frac{1}{2} + B \cdot \frac{\sqrt{3}}{2} = 2 \quad (1) \] \[ -A \cdot \frac{\sqrt{3}}{2} + B \cdot \frac{1}{2} = -4 \quad (2) \] --- ### Step 4: Solve for \(A\) and \(B\) Multiply both equations by 2 to clear denominators: \[ A + B \sqrt{3} = 4 \quad (3) \] \[ -A \sqrt{3} + B = -8 \quad (4) \] From (3): \[ A = 4 - B \sqrt{3} \] Substitute into (4): \[ - (4 - B \sqrt{3}) \sqrt{3} + B = -8 \] \[ -4 \sqrt{3} + B \sqrt{3} \cdot \sqrt{3} + B = -8 \] \[ -4 \sqrt{3} + B \cdot 3 + B = -8 \] \[ -4 \sqrt{3} + 4 B = -8 \] Solve for \( B \): \[ 4 B = -8 + 4 \sqrt{3} \] \[ B = -2 + \sqrt{3} \] Now, find \( A \): \[ A = 4 - B \sqrt{3} = 4 - (-2 + \sqrt{3}) \sqrt{3} \] \[ A = 4 + 2 \sqrt{3} - (\sqrt{3} \times \sqrt{3}) = 4 + 2 \sqrt{3} - 3 = 1 + 2 \sqrt{3} \] --- ### **Final solution for (a):** \[ \boxed{ \boxed{ \boxed{ y(t) = (1 + 2 \sqrt{3}) \cos t + (-2 + \sqrt{3}) \sin t } } } \] --- ## (b) \( y'' - 2 y' + 5 y = 0 \), with \( y(\pi/2) = 0 \), \( y'(\pi/2) = 2 \) ### Step 1: Find the characteristic roots \[ r^2 - 2 r + 5 = 0 \] Discriminant: \[ \Delta = (-2)^2 - 4 \times 1 \times 5 = 4 - 20 = -16 \] Roots: \[ r = \frac{2 \pm \sqrt{-16}}{2} = 1 \pm 2 i \] ### Step 2: Write the general solution \[ \boxed{ y(t) = e^{t} \left( C_1 \cos 2t + C_2 \sin 2t \right) \] ### Step 3: Find the derivative \( y'(t) \) Using product rule: \[ y'(t) = e^{t}( C_1 \cos 2t + C_2 \sin 2t ) + e^{t} \left( -2 C_1 \sin 2t + 2 C_2 \cos 2t \right) \] \[ = e^{t} \left[ C_1 \cos 2t + C_2 \sin 2t - 2 C_1 \sin 2t + 2 C_2 \cos 2t \right] \] Group similar terms: \[ y'(t) = e^{t} \left[ ( C_1 + 2 C_2 ) \cos 2t + ( C_2 - 2 C_1 ) \sin 2t \right] \] --- ### Step 4: Apply initial conditions at \( t = \pi/2 \): - \( y(\pi/2) = 0 \): \[ e^{\pi/2} \left[ C_1 \cos \pi + C_2 \sin \pi \right] = 0 \] \[ e^{\pi/2} \left[ C_1 \times (-1) + C_2 \times 0 \right] = 0 \] \[ - C_1 e^{\pi/2} = 0 \Rightarrow C_1 = 0 \] - \( y'(\pi/2) = 2 \): \[ e^{\pi/2} \left[ ( 0 + 2 C_2 ) \cos \pi + ( C_2 - 0 ) \sin \pi \right] = 2 \] \[ e^{\pi/2} \left[ 2 C_2 \times (-1) + C_2 \times 0 \right] = 2 \] \[ -2 C_2 e^{\pi/2} = 2 \] \[ C_2 = - \frac{1}{e^{\pi/2}} \] --- ### **Final solution for (b):** \[ \boxed{ \boxed{ y(t) = e^{t} \left( 0 \times \cos 2t + \left( - \frac{1}{e^{\pi/2}} \right) \sin 2t \right) = - \frac{e^{t}}{e^{\pi/2}} \sin 2t } } \] or simplified: \[ \boxed{ \boxed{ y(t) = - e^{t - \pi/2} \sin 2t } } \] --- ## (c) \( y''' + 12 y'' + 36 y' = 0 \), with \( y(0) = 0 \), \( y'(0) = 1 \), \( y''(0) = 6 \) ### Step 1: Find the roots of the characteristic equation \[ r^3 + 12 r^2 + 36 r = 0 \] Factor: \[ r(r^2 + 12 r + 36) = 0 \] Quadratic: \[ r^2 + 12 r + 36 = 0 \] Discriminant: \[ \Delta = 144 - 144 = 0 \] Roots: \[ r = -6 \quad \text{(double root)} \] and \[ r = 0 \] ### Step 2: General solution Since roots are \( r = 0, -6, -6 \), the general solution: \[ \boxed{ y(t) = A + B e^{-6 t} + C t e^{-6 t} \] --- ### Step 3: Derivatives - \( y(t) = A + B e^{-6 t} + C t e^{-6 t} \) Compute \( y'(t) \): \[ y'(t) = 0 + B (-6) e^{-6 t} + C \left( e^{-6 t} - 6 t e^{-6 t} \right) \] \[ = -6 B e^{-6 t} + C e^{-6 t} - 6 C t e^{-6 t} \] \[ = \left( -6 B + C \right) e^{-6 t} - 6 C t e^{-6 t} \] Compute \( y''(t) \): \[ \frac{d}{dt} y'(t) = \left( -6 B + C \right) (-6) e^{-6 t} - 6 C \left( e^{-6 t} - 6 t e^{-6 t} \right) \] \[ = 36 B e^{-6 t} - 6 C e^{-6 t} + 36 C t e^{-6 t} \] ### Step 4: Apply initial conditions at \( t=0 \): - \( y(0) = A + B + 0 = 0 \Rightarrow A = - B \) - \( y'(0) = \left( -6 B + C \right) - 0 = 1 \Rightarrow -6 B + C = 1 \) - \( y''(0) = 36 B - 6 C + 0 = 6 \Rightarrow 36 B - 6 C = 6 \) Now solve for \( B \) and \( C \): From \( -6 B + C = 1 \): \[ C = 1 + 6 B \] From \( 36 B - 6 C = 6 \): \[ 36 B - 6 ( 1 + 6 B ) = 6 \] \[ 36 B - 6 - 36 B = 6 \] \[ -6 = 6 \] This is a contradiction, indicating an inconsistency. **But wait!** Let's carefully re-check the derivatives. Our earlier derivative calculations are approximate; perhaps a more systematic approach is better. --- ### **Alternative: Use derivatives explicitly** Express \( y(t) \): \[ y(t) = A + B e^{-6 t} + C t e^{-6 t} \] Calculate derivatives: - \( y' \): \[ y' = B (-6) e^{-6 t} + C \left( e^{-6 t} - 6 t e^{-6 t} \right) \] \[ = -6 B e^{-6 t} + C e^{-6 t} - 6 C t e^{-6 t} \] \[ = \left( -6 B + C \right) e^{-6 t} - 6 C t e^{-6 t} \] - \( y'' \): \[ \frac{d}{dt} y' = \left( -6 B + C \right) (-6) e^{-6 t} - 6 C \left( e^{-6 t} - 6 t e^{-6 t} \right) \] \[ = 36 B e^{-6 t} - 6 C e^{-6 t} - 6 C e^{-6 t} + 36 C t e^{-6 t} \] \[ = 36 B e^{-6 t} - 12 C e^{-6 t} + 36 C t e^{-6 t} \] --- ### **Applying initial conditions:** At \( t=0 \): \[ y(0) = A + B e^{0} + C \times 0 \times e^{0} = A + B = 0 \Rightarrow A = -B \] \[ y'(0) = \left( -6 B + C \right) e^{0} - 6 C \times 0 \times e^{0} = -6 B + C = 1 \] \[ y''(0) = 36 B e^{0} - 12 C e^{0} + 36 C \times 0 \times e^{0} = 36 B - 12 C = 6 \] Now, from the second equation: \[ -6 B + C = 1 \Rightarrow C = 1 + 6 B \] From the third: \[ 36 B - 12 C = 6 \] Substitute \( C = 1 + 6 B \): \[ 36 B - 12 (1 + 6 B) = 6 \] \[ 36 B - 12 - 72 B = 6 \] \[ (36 B - 72 B) - 12 = 6 \] \[ -36 B - 12 = 6 \] \[ -36 B = 18 \] \[ B = -\frac{18}{36} = -\frac{1}{2} \] Now compute \( C \): \[ C = 1 + 6 \times (-\frac{1}{2}) = 1 - 3 = -2 \] Recall \( A = - B \): \[ A = - (-\frac{1}{2}) = \frac{1}{2} \] --- ### **Final solution:** \[ \boxed{ \boxed{ y(t) = \frac{1}{2} + \left(-\frac{1}{2}\right) e^{-6

Apologies for the interruption. Continuing from where we left off: --- ### Final solution for (c): \[ \boxed{ y(t) = \frac{1}{2} - \frac{1}{2} e^{-6 t} - 2 t e^{-6 t} } \] --- ## **Summary of all solutions:** ### (a) \[ \boxed{ y(t) = (1 + 2 \sqrt{3}) \cos t + (-2 + \sqrt{3}) \sin t } \] ### (b) \[ \boxed{ y(t) = - e^{t - \pi/2} \sin 2t } \] ### (c) \[ \boxed{ y(t) = \frac{1}{2} - \frac{1}{2} e^{-6 t} - 2 t e^{-6 t} } \] --- ## **Behavior for increasing \( t \):** - **(a)**: The solution is a combination of sine and cosine functions with constant amplitude, representing **oscillations** that persist indefinitely. - **(b)**: The solution involves an exponential growth factor \( e^{t} \) multiplied by a sine function, indicating **oscillations with amplitudes that grow exponentially** as \( t \to \infty \). - **(c)**: The solution approaches a **steady state** \( y(t) \to \frac{1}{2} \) as \( t \to \infty \), since the exponential terms decay to zero. The solution **smoothly converges** to this constant. --- ## **Sketching the graphs:** - For **(a)**: The graph is a bounded oscillation around zero. - For **(b)**: The oscillations grow larger with increasing \( t \). - For **(c)**: The graph starts at 0 (since \( y(0) = 0 \)) and approaches \( \frac{1}{2} \) as \( t \to \infty \). --- If you need plots or further explanation, please let me know!