Let's solve the initial value problem step by step: \[ y'' + (\cos t)y' + (\sin t)y = 0, \quad y(0) = 0, \quad y'(0) = 0 \] --- ## Step 1: Write the equation \[ y'' + (\cos t)y' + (\sin t)y = 0 \] --- ## Step 2: Try a power series solution Assume \( y(t) = \sum_{n=0}^\infty a_n t^n \). Then, \[ y'(t) = \sum_{n=1}^\infty n a_n t^{n-1} \] \[ y''(t) = \sum_{n=2}^\infty n(n-1) a_n t^{n-2} \] --- ## Step 3: Initial conditions From \( y(0) = 0 \), \( a_0 = 0 \). From \( y'(0) = 0 \), \( y'(0) = a_1 \), so \( a_1 = 0 \). --- ## Step 4: Substitute series into the ODE Write \(\cos t\) and \(\sin t\) as series: \[ \cos t = 1 - \frac{t^2}{2!} + \frac{t^4}{4!} - \cdots \] \[ \sin t = t - \frac{t^3}{3!} + \frac{t^5}{5!} - \cdots \] So, \[ (\cos t)y' = \left(1 - \frac{t^2}{2} + \cdots\right)\left(\sum_{n=1}^\infty n a_n t^{n-1}\right) \] \[ (\sin t)y = \left(t - \frac{t^3}{6} + \cdots\right)\left(\sum_{n=0}^\infty a_n t^n\right) \] Let's write out the first few terms for each: #### \(y''\): \[ y'' = 2 \cdot 1 a_2 + 3 \cdot 2 a_3 t + 4 \cdot 3 a_4 t^2 + \cdots \] #### \((\cos t)y'\): \[ y' = a_1 + 2a_2 t + 3a_3 t^2 + 4a_4 t^3 + \cdots \] \[ (\cos t)y' = (a_1 + 2a_2 t + 3a_3 t^2 + 4a_4 t^3 + \cdots) - \frac{t^2}{2}(a_1 + 2a_2 t + 3a_3 t^2 + \cdots) + \cdots \] #### \((\sin t)y\): \[ y = a_0 + a_1 t + a_2 t^2 + a_3 t^3 + a_4 t^4 + \cdots \] \[ (\sin t)y = t(a_0 + a_1 t + a_2 t^2 + \cdots) - \frac{t^3}{6}(a_0 + a_1 t + \cdots) + \cdots \] --- ## Step 5: Collect terms of equal powers (up to \(t^3\)) Set up the recurrence for coefficients. ### \(t^0\) term: - Only \(y''\) and \((\cos t)y'\) contribute: \[ y'': 2a_2 \] \[ (\cos t)y': a_1 \] \[ (\sin t)y: 0 \] So: \[ 2a_2 + a_1 = 0 \implies a_2 = 0 \text{ (since \(a_1 = 0\))} \] ### \(t^1\) term: \[ y'': 3 \cdot 2 a_3 t \] \[ (\cos t)y': 2a_2 t \] \[ (\sin t)y: a_0 t \] So: \[ 6a_3 + 2a_2 + a_0 = 0 \implies 6a_3 = 0 \implies a_3 = 0 \] ### \(t^2\) term: \[ y'': 4 \cdot 3 a_4 t^2 = 12 a_4 \] \[ (\cos t)y': 3a_3 t^2 - \frac{a_1}{2} t^2 \] \[ (\sin t)y: a_1 t^2 \] So: \[ 12a_4 + 3a_3 - \frac{a_1}{2} + a_1 = 0 \] But \(a_1 = a_3 = 0\): \[ 12a_4 = 0 \implies a_4 = 0 \] ### \(t^3\) term: \[ y'': 5 \cdot 4 a_5 t^3 = 20 a_5 \] \[ (\cos t)y': 4a_4 t^3 - a_2 t^3 \] \[ (\sin t)y: a_2 t^3 - \frac{a_0}{6} t^3 \] So: \[ 20a_5 + 4a_4 - a_2 + a_2 - \frac{a_0}{6} = 0 \implies 20a_5 + 4a_4 - \frac{a_0}{6} = 0 \] But \(a_4 = a_2 = a_0 = 0\): \[ 20a_5 = 0 \implies a_5 = 0 \] --- ## Step 6: Pattern All coefficients \(a_n = 0\). So the **only solution** is: \[ \boxed{y(t) = 0} \] --- ## Step 7: Final Answer \[ \boxed{y(t) = 0} \] This is the only solution that satisfies both the equation and the initial conditions.

Let's carefully analyze and solve the initial value problem step by step with detailed calculations. --- ### Given differential equation: \[ y'' + (\cos t) y' + (\sin t) y = 0, \] with initial conditions: \[ y(0) = 0, \quad y'(0) = 0. \] --- ### **Step 1: Recognize the nature of the problem** Since the coefficients involve trigonometric functions, and the initial conditions are at \(t=0\), a power series solution centered at \(t=0\) is suitable. --- ### **Step 2: Assume a power series solution** Suppose: \[ y(t) = \sum_{n=0}^\infty a_n t^n. \] Then: \[ y'(t) = \sum_{n=1}^\infty n a_n t^{n-1}, \] \[ y''(t) = \sum_{n=2}^\infty n(n-1) a_n t^{n-2}. \] --- ### **Step 3: Use initial conditions to find some coefficients** At \(t=0\): \[ y(0) = a_0 = 0, \] \[ y'(0) = a_1 = 0. \] --- ### **Step 4: Express \(\cos t\) and \(\sin t\) as power series** \[ \cos t = 1 - \frac{t^2}{2} + \frac{t^4}{24} - \frac{t^6}{720} + \cdots, \] \[ \sin t = t - \frac{t^3}{6} + \frac{t^5}{120} - \frac{t^7}{5040} + \cdots. \] --- ### **Step 5: Write \( (\cos t)y' \) and \( (\sin t) y \) as series** \[ (\cos t) y' = \left(1 - \frac{t^2}{2} + \frac{t^4}{24} - \cdots \right) \left( \sum_{n=1}^\infty n a_n t^{n-1} \right), \] \[ (\sin t) y = \left(t - \frac{t^3}{6} + \frac{t^5}{120} - \cdots \right) \left( \sum_{n=0}^\infty a_n t^n \right). \] --- ### **Step 6: Find the coefficients \(a_n\) by substituting into the differential equation** Our goal is to find a recurrence relation for \(a_n\). Let's proceed term-by-term. --- ### **Step 7: Express derivatives** \[ y'' = \sum_{n=2}^\infty n(n-1) a_n t^{n-2}. \] --- ### **Step 8: Write the entire differential equation in series form** The equation: \[ y'' + (\cos t) y' + (\sin t) y = 0, \] becomes: \[ \sum_{n=2}^\infty n(n-1) a_n t^{n-2} + \left( \sum_{k=0}^\infty c_k t^k \right) \left( \sum_{m=1}^\infty m a_m t^{m-1} \right) + \left( \sum_{p=0}^\infty s_p t^p \right) \left( \sum_{q=0}^\infty a_q t^q \right) = 0, \] where \(c_k\) are coefficients from \(\cos t\), and \(s_p\) from \(\sin t\). --- ### **Step 9: Find the first few terms explicitly** We will compute the first few terms to find the pattern. --- ### **Step 10: Initial terms** - \(a_0 = 0\), - \(a_1 = 0\), - \(a_2\), \(a_3\), etc., are unknowns to be determined. --- ### **Step 11: Express series for \((\cos t) y'\)** \[ (\cos t) y' = \left(1 - \frac{t^2}{2} + \frac{t^4}{24} - \cdots \right) \left( a_1 + 2 a_2 t + 3 a_3 t^2 + 4 a_4 t^3 + \cdots \right). \] Since \(a_1=0\), this simplifies to: \[ (\cos t) y' = \left(1 - \frac{t^2}{2} + \frac{t^4}{24} - \cdots \right) \left( 2 a_2 t + 3 a_3 t^2 + 4 a_4 t^3 + \cdots \right). \] Calculating the first few terms: - For the \(t\) term: \[ 1 \times 2 a_2 t = 2 a_2 t, \] - For the \(t^2\) term: \[ 1 \times 3 a_3 t^2 - \frac{t^2}{2} \times 2 a_2 t = 3 a_3 t^2 - a_2 t^2, \] - For the \(t^3\) term: \[ 1 \times 4 a_4 t^3 - \frac{t^2}{2} \times 3 a_3 t^2 + \frac{t^4}{24} \times 2 a_2 t, \] which simplifies to: \[ 4 a_4 t^3 - \frac{3 a_3}{2} t^4 + \text{higher order terms}. \] Similarly, for the initial terms: \[ (\cos t) y' \approx 2 a_2 t + (3 a_3 - a_2) t^2 + \text{higher order terms}. \] --- ### **Step 12: Express series for \((\sin t) y\)** \[ (\sin t) y = \left( t - \frac{t^3}{6} + \frac{t^5}{120} - \cdots \right) \left( a_0 + a_1 t + a_2 t^2 + a_3 t^3 + \cdots \right). \] Since \(a_0=0\), \(a_1=0\), this reduces to: \[ (\sin t) y = t (a_2 t^2 + a_3 t^3 + a_4 t^4 + \cdots) - \frac{t^3}{6} (a_2 t^2 + a_3 t^3 + \cdots). \] Calculating: - For \(t^3\): \[ a_2 t^3, \] - For \(t^4\): \[ a_3 t^4 - \frac{a_2}{6} t^4. \] Higher order terms involve higher powers. --- ### **Step 13: Write the full series and equate coefficients** Now, the original differential equation reduces to: \[ \sum_{n=2}^\infty n(n-1) a_n t^{n-2} + \left( \text{series for } (\cos t) y' \right) + \left( \text{series for } (\sin t) y \right) = 0. \] At each power of \(t\), the sum of coefficients must be zero. --- ### **Step 14: Collect terms for specific powers** Let's analyze terms at \(t^0\), \(t^1\), \(t^2\), etc. --- ### **At \(t^0\):** - \(y''\) contributes nothing (since \(n \geq 2\), so \(n-2 \geq 0\)): \[ \text{from } y'': \text{coefficient at } t^0: 2 a_2 \text{ (from } n=2) \text{ with } t^{0}\text{? No, } t^{n-2} \Rightarrow n=2: t^{0} \text{ term}: \] \[ \text{coefficient} = 2 \times 1 \times a_2 = 2 a_2. \] - \( (\cos t) y'\): \[ \text{coefficient at } t^0: 0, \] since its expansion starts from \(t^1\). - \( (\sin t) y \): \[ a_2 t^3 \text{ contributes at } t^3, \] so no contribution at \(t^0\). Therefore, **equation at \(t^0\):** \[ 2 a_2 = 0 \implies a_2 = 0. \] --- ### **At \(t^1\):** - \( y'' \): \[ n=3 \Rightarrow 3 \times 2 \times a_3 = 6 a_3, \] but \(t^{1}\) corresponds to \(n=3\), which gives \(t^{1}\) term. - \( (\cos t) y' \): \[ 2 a_2 t, \] but \(a_2=0\), so no contribution. - \( (\sin t) y \): \[ a_2 t^3 \text{, no contribution at } t^1, \] Thus, at \(t^1\): \[ 6 a_3 = 0 \implies a_3=0. \] --- ### **At \(t^2\):** - \( y'' \): \[ n=4: 4 \times 3 \times a_4 = 12 a_4, \] contribution at \(t^2\). - \( (\cos t) y' \): \[ 3 a_3 t^2 - \frac{a_2}{2} t^{2}, \] but \(a_2=0\), \(a_3=0\), so total zero. - \( (\sin t) y \): \[ a_2 t^3 \text{, no contribution at } t^2. \] So at \(t^2\): \[ 12 a_4 = 0 \implies a_4=0. \] --- ### **At \(t^3\):** - \( y'' \): \[ n=5: 5 \times 4 \times a_5 = 20 a_5, \] contribution at \(t^3\). - \( (\cos t) y' \): \[ 4 a_4 t^3 - \frac{a_3}{2} t^3, \] but \(a_3=0, a_4=0\). - \( (\sin t) y \): \[ a_2 t^3, \] but \(a_2=0\). At \(t^3\): \[ 20 a_5 = 0 \implies a_5=0. \] --- ### **Conclusion of recursion:** All coefficients \(a_n=0\) for \(n \ge 1\). Since initial conditions give \(a_0=0\) and \(a_1=0\), the **only solution** is: \[ \boxed{ \boxed{ y(t) = 0 } } \] --- ### **Final answer:** \[ \boxed{ \boxed{ y(t) = 0 } } \] This trivial solution satisfies the differential equation and initial conditions. --- ## **Summary:** - Assumed power series solution. - Used initial conditions to determine \(a_0=0\), \(a_1=0\). - Calculated successive coefficients via series substitution. - Found that all coefficients \(a_n=0\), so the only solution is the zero function. --- **Note:** This indicates the differential equation leads to the trivial solution under the prescribed initial conditions.