Let's solve the system using **Laplace transforms**. Given: \[ \begin{cases} x' - 4x + 4y = \sin t, & x(0) = 0 \\ \frac{13}{4}x - y' - 2y = \cos t, & y(0) = 0 \end{cases} \] Let \( X(s) = \mathcal{L}[x(t)] \), \( Y(s) = \mathcal{L}[y(t)] \). ### Step 1: Take Laplace Transforms Recall: - \( \mathcal{L}[x'] = sX(s) - x(0) \) - \( \mathcal{L}[y'] = sY(s) - y(0) \) - \( \mathcal{L}[\sin t] = \frac{1}{s^2+1} \) - \( \mathcal{L}[\cos t] = \frac{s}{s^2+1} \) #### 1st Equation: \[ x' - 4x + 4y = \sin t \] \[ s X(s) - x(0) - 4 X(s) + 4 Y(s) = \frac{1}{s^2+1} \] \[ sX(s) - 4X(s) + 4Y(s) = \frac{1}{s^2+1} \] \[ (s-4)X(s) + 4Y(s) = \frac{1}{s^2+1} \] #### 2nd Equation: \[ \frac{13}{4}x - y' - 2y = \cos t \] \[ \frac{13}{4} X(s) - (s Y(s) - y(0)) - 2 Y(s) = \frac{s}{s^2+1} \] \[ \frac{13}{4} X(s) - s Y(s) - 2 Y(s) = \frac{s}{s^2+1} \] \[ \frac{13}{4} X(s) - (s + 2) Y(s) = \frac{s}{s^2+1} \] --- ### Step 2: Solve for \( X(s) \) and \( Y(s) \) You have the system: \[ \begin{cases} (s-4)X(s) + 4Y(s) = \frac{1}{s^2+1} \quad \cdots (1) \\ \frac{13}{4}X(s) - (s+2)Y(s) = \frac{s}{s^2+1} \quad \cdots (2) \end{cases} \] Let's solve for \( X(s) \) and \( Y(s) \): #### Multiply (2) by 4 to clear fractions: \[ 13 X(s) - 4(s+2)Y(s) = \frac{4s}{s^2+1} \] Now, let's solve the system: **From (1):** \[ (s-4)X(s) + 4Y(s) = \frac{1}{s^2+1} \] \[ 4Y(s) = \frac{1}{s^2+1} - (s-4)X(s) \] \[ Y(s) = \frac{1}{4(s^2+1)} - \frac{(s-4)}{4}X(s) \] Substitute \( Y(s) \) into the second equation: **Put into the above:** \[ 13X(s) - 4(s+2)\left[\frac{1}{4(s^2+1)} - \frac{(s-4)}{4}X(s)\right] = \frac{4s}{s^2+1} \] \[ 13X(s) - (s+2)\left[\frac{1}{s^2+1} - (s-4)X(s)\right] = \frac{4s}{s^2+1} \] \[ 13X(s) - \frac{s+2}{s^2+1} + (s+2)(s-4)X(s) = \frac{4s}{s^2+1} \] \[ \left[13 + (s+2)(s-4)\right]X(s) = \frac{4s}{s^2+1} + \frac{s+2}{s^2+1} \] \[ \left[13 + (s^2-4s+2s-8)\right]X(s) = \frac{4s+s+2}{s^2+1} \] \[ \left[13 + s^2 - 2s - 8\right]X(s) = \frac{5s+2}{s^2+1} \] \[ (s^2 - 2s + 5)X(s) = \frac{5s+2}{s^2+1} \] \[ X(s) = \frac{5s+2}{(s^2 - 2s + 5)(s^2+1)} \] --- Now, substitute back for \( Y(s) \): \[ Y(s) = \frac{1}{4(s^2+1)} - \frac{(s-4)}{4}X(s) \] \[ Y(s) = \frac{1}{4(s^2+1)} - \frac{s-4}{4} \cdot \frac{5s+2}{(s^2 - 2s + 5)(s^2+1)} \] --- ### Step 3: Final Answers (in terms of Laplace transforms) \[ \boxed{ \begin{align*} X(s) &= \frac{5s+2}{(s^2-2s+5)(s^2+1)} \\ Y(s) &= \frac{1}{4(s^2+1)} - \frac{s-4}{4}\cdot\frac{5s+2}{(s^2-2s+5)(s^2+1)} \end{align*} } \] ### If you want the time-domain solution \( x(t) \), \( y(t) \): You can leave the answer in terms of Laplace transforms as above (as the question allows). If you want to invert them to \(x(t)\) and \(y(t)\), let me know, and I can proceed further with partial fractions and inverse Laplace!

Certainly! Let's go through the detailed step-by-step solution using Laplace transforms for the given system. --- ## **Given system:** \[ \begin{cases} x' - 4x + 4y = \sin t, & x(0) = 0 \\ \frac{13}{4}x - y' - 2y = \cos t, & y(0) = 0 \end{cases} \] --- ## **Step 1: Take Laplace Transforms** Recall: - \( \mathcal{L}[x'(t)] = sX(s) - x(0) \) - \( \mathcal{L}[y'(t)] = sY(s) - y(0) \) - \( \mathcal{L}[\sin t] = \frac{1}{s^2 + 1} \) - \( \mathcal{L}[\cos t] = \frac{s}{s^2 + 1} \) Given initial conditions \( x(0)=0, y(0)=0 \), the transforms are straightforward. --- ## **Step 2: Transform each equation** ### First equation: \[ x' - 4x + 4y = \sin t \] Laplace transform: \[ sX(s) - 0 - 4X(s) + 4Y(s) = \frac{1}{s^2 + 1} \] Simplify: \[ (s - 4)X(s) + 4Y(s) = \frac{1}{s^2 + 1} \quad \text{(Equation 1)} \] ### Second equation: \[ \frac{13}{4}x - y' - 2y = \cos t \] Laplace transform: \[ \frac{13}{4}X(s) - [sY(s) - 0] - 2Y(s) = \frac{s}{s^2 + 1} \] Simplify: \[ \frac{13}{4}X(s) - sY(s) - 2Y(s) = \frac{s}{s^2 + 1} \] \[ \frac{13}{4}X(s) - (s + 2)Y(s) = \frac{s}{s^2 + 1} \quad \text{(Equation 2)} \] --- ## **Step 3: Solve the system for \(X(s)\) and \(Y(s)\)** **Equation 1:** \[ (s - 4)X(s) + 4Y(s) = \frac{1}{s^2 + 1} \] **Equation 2:** \[ \frac{13}{4}X(s) - (s + 2)Y(s) = \frac{s}{s^2 + 1} \] --- ## **Step 4: Express \(Y(s)\) from Equation 1** \[ 4Y(s) = \frac{1}{s^2 + 1} - (s - 4)X(s) \] \[ Y(s) = \frac{1}{4(s^2 + 1)} - \frac{s - 4}{4} X(s) \] --- ## **Step 5: Substitute \(Y(s)\) into Equation 2** Equation 2: \[ \frac{13}{4}X(s) - (s + 2)Y(s) = \frac{s}{s^2 + 1} \] Substitute \(Y(s)\): \[ \frac{13}{4}X(s) - (s + 2) \left[ \frac{1}{4(s^2 + 1)} - \frac{s - 4}{4}X(s) \right] = \frac{s}{s^2 + 1} \] Distribute: \[ \frac{13}{4}X(s) - \frac{s + 2}{4(s^2 + 1)} + \frac{s + 2}{4} (s - 4) X(s) = \frac{s}{s^2 + 1} \] Let's write all terms involving \(X(s)\): \[ \left[ \frac{13}{4} + \frac{s + 2}{4}(s - 4) \right] X(s) - \frac{s + 2}{4(s^2 + 1)} = \frac{s}{s^2 + 1} \] --- ## **Step 6: Simplify the coefficient of \(X(s)\)** Calculate: \[ \frac{s + 2}{4}(s - 4) = \frac{1}{4}(s + 2)(s - 4) \] Expand numerator: \[ (s + 2)(s - 4) = s^2 - 4s + 2s -8 = s^2 - 2s -8 \] So: \[ \frac{1}{4}(s^2 - 2s -8) = \frac{s^2}{4} - \frac{2s}{4} - 2 = \frac{s^2}{4} - \frac{s}{2} - 2 \] Now, the coefficient of \(X(s)\): \[ \frac{13}{4} + \left(\frac{s^2}{4} - \frac{s}{2} - 2 \right) = \frac{13}{4} + \frac{s^2}{4} - \frac{s}{2} - 2 \] Express all with common denominator 4: \[ \frac{13}{4} + \frac{s^2}{4} - \frac{2s}{4} - \frac{8}{4} = \frac{13 + s^2 - 2s - 8}{4} = \frac{s^2 - 2s + 5}{4} \] Thus, the entire equation becomes: \[ \frac{s^2 - 2s + 5}{4} X(s) - \frac{s + 2}{4(s^2 + 1)} = \frac{s}{s^2 + 1} \] --- ## **Step 7: Isolate \(X(s)\)** Bring the second term to RHS: \[ \frac{s^2 - 2s + 5}{4} X(s) = \frac{s}{s^2 + 1} + \frac{s + 2}{4(s^2 + 1)} \] Express RHS with common denominator \(4(s^2 + 1)\): \[ \frac{4s}{4(s^2 + 1)} + \frac{s + 2}{4(s^2 + 1)} = \frac{4s + s + 2}{4(s^2 + 1)} = \frac{5s + 2}{4(s^2 + 1)} \] Now: \[ X(s) = \frac{4}{s^2 - 2s + 5} \times \frac{5s + 2}{4(s^2 + 1)} = \frac{5s + 2}{(s^2 - 2s + 5)(s^2 + 1)} \] **Final expression for \(X(s)\):** \[ \boxed{ X(s) = \frac{5s + 2}{(s^2 - 2s + 5)(s^2 + 1)} } \] --- ## **Step 8: Find \(Y(s)\)** Recall: \[ Y(s) = \frac{1}{4(s^2 + 1)} - \frac{s - 4}{4} X(s) \] Substitute \(X(s)\): \[ Y(s) = \frac{1}{4(s^2 + 1)} - \frac{s - 4}{4} \times \frac{5s + 2}{(s^2 - 2s + 5)(s^2 + 1)} \] --- ## **Final answer:** \[ \boxed{ \begin{aligned} X(s) &= \frac{5s + 2}{(s^2 - 2s + 5)(s^2 + 1)} \\ Y(s) &= \frac{1}{4(s^2 + 1)} - \frac{s - 4}{4} \times \frac{5s + 2}{(s^2 - 2s + 5)(s^2 + 1)} \end{aligned} } \] --- ## **Summary:** - We transformed the original differential equations into algebraic equations using Laplace transforms. - Solved the algebraic system for \(X(s)\) and \(Y(s)\). - Expressed solutions explicitly in \(s\). - For the inverse transform (if desired), partial fraction decomposition would be necessary to find \(x(t)\) and \(y(t)\). --- **Let me know if you'd like me to proceed further with inverse Laplace transforms!**

Certainly! Let's proceed to find the inverse Laplace transforms of \(X(s)\) and \(Y(s)\) in terms of exponentials \(e^{at}\), which typically appear in inverse transforms of quadratic denominator factors. --- ## **Step 9: Complete the inverse Laplace transform of \(X(s)\)** Recall: \[ X(s) = \frac{5s + 2}{(s^2 - 2s + 5)(s^2 + 1)} \] ### Step 9.1: Rewrite denominators in completed square form: \[ s^2 - 2s + 5 = (s - 1)^2 + 4 \] \[ s^2 + 1 = s^2 + 1 \] ### Step 9.2: Partial fraction decomposition We want to express: \[ X(s) = \frac{5s + 2}{[(s - 1)^2 + 4](s^2 + 1)} \] Note that: \[ s^2 + 1 = (s)^2 + 1 \] Our goal is to write: \[ \frac{5s + 2}{[(s - 1)^2 + 4](s^2 + 1)} = \frac{A s + B}{(s - 1)^2 + 4} + \frac{C s + D}{s^2 + 1} \] Multiply both sides by the denominator: \[ 5s + 2 = (A s + B)(s^2 + 1) + (C s + D)((s - 1)^2 + 4) \] --- ## **Step 10: Expand RHS** \[ (A s + B)(s^2 + 1) = A s (s^2 + 1) + B (s^2 + 1) = A s^3 + A s + B s^2 + B \] \[ (C s + D)((s - 1)^2 + 4) = (C s + D)(s^2 - 2s + 1 + 4) = (C s + D)(s^2 - 2s + 5) \] Now expand: \[ C s (s^2 - 2s + 5) + D (s^2 - 2s + 5) = C s^3 - 2 C s^2 + 5 C s + D s^2 - 2 D s + 5 D \] Combine all parts: \[ \text{RHS} = A s^3 + A s + B s^2 + B + C s^3 - 2 C s^2 + 5 C s + D s^2 - 2 D s + 5 D \] Group like terms: - \(s^3\): \(A + C\) - \(s^2\): \(B - 2 C + D\) - \(s\): \(A + 5 C - 2 D\) - constants: \(B + 5 D\) Set equal to LHS: \[ 5s + 2 \] So, \[ \boxed{ \begin{cases} A + C = 0 \\ B - 2 C + D = 0 \\ A + 5 C - 2 D = 5 \\ B + 5 D = 2 \end{cases} } \] --- ## **Step 11: Solve for coefficients** From the first eq: \[ A + C = 0 \Rightarrow C = -A \] From the fourth eq: \[ B + 5 D = 2 \Rightarrow B = 2 - 5 D \] Substitute \(C = -A\) into the third eq: \[ A + 5(-A) - 2 D = 5 \] \[ A - 5A - 2 D = 5 \] \[ -4A - 2 D = 5 \] \[ 2A + D = -\frac{5}{2} \] \[ D = -\frac{5}{2} - 2A \] Now substitute \(B = 2 - 5 D\) and \(D\) into the second eq: \[ B - 2 C + D = 0 \] \[ (2 - 5 D) - 2(-A) + D = 0 \] \[ 2 - 5 D + 2A + D = 0 \] \[ 2 + 2A - 4 D = 0 \] But \(D = -\frac{5}{2} - 2A\), so: \[ 2 + 2A - 4 \left(-\frac{5}{2} - 2A\right) = 0 \] Calculate: \[ 2 + 2A + 4 \times \frac{5}{2} + 8A = 0 \] \[ 2 + 2A + 10 + 8A = 0 \] \[ (2 + 10) + (2A + 8A) = 0 \] \[ 12 + 10A = 0 \] \[ 10A = -12 \] \[ A = -\frac{6}{5} \] Now \(D = -\frac{5}{2} - 2A\): \[ D = -\frac{5}{2} - 2 \times \left(-\frac{6}{5}\right) = -\frac{5}{2} + \frac{12}{5} \] Express with common denominator 10: \[ -\frac{25}{10} + \frac{24}{10} = -\frac{1}{10} \] Recall \(C = -A\): \[ C = -(-\frac{6}{5}) = \frac{6}{5} \] And \(B = 2 - 5 D\): \[ B = 2 - 5 \times \left(-\frac{1}{10}\right) = 2 + \frac{5}{10} = 2 + \frac{1}{2} = \frac{5}{2} \] --- ## **Step 12: Final partial fraction form** \[ X(s) = \frac{A s + B}{(s - 1)^2 + 4} + \frac{C s + D}{s^2 + 1} \] \[ X(s) = \frac{-\frac{6}{5} s + \frac{5}{2}}{(s - 1)^2 + 4} + \frac{\frac{6}{5} s - \frac{1}{10}}{s^2 + 1} \] --- ## **Step 13: Inverse Laplace transforms** Recall standard transforms: \[ \mathcal{L}^{-1}\left[\frac{s - a}{(s - a)^2 + b^2}\right] = e^{a t} \cos (b t) \] \[ \mathcal{L}^{-1}\left[\frac{b}{(s - a)^2 + b^2}\right] = e^{a t} \sin (b t) \] Express numerator terms in these forms: ### For first term: \[ -\frac{6}{5} s + \frac{5}{2} \] Write as: \[ -\frac{6}{5}(s - 1 + 1) + \frac{5}{2} = -\frac{6}{5}(s - 1) - \frac{6}{5} + \frac{5}{2} \] So, \[ X(s) = \frac{-\frac{6}{5}(s - 1) + \left( - \frac{6}{5} + \frac{5}{2} \right)}{(s - 1)^2 + 4} + \frac{\frac{6}{5} s - \frac{1}{10}}{s^2 + 1} \] The constants: \[ - \frac{6}{5} + \frac{5}{2} = -\frac{6}{5} + \frac{25}{10} = -\frac{12}{10} + \frac{25}{10} = \frac{13}{10} \] Similarly, write second numerator: \[ \frac{6}{5} s - \frac{1}{10} = \frac{6}{5} (s) - \frac{1}{10} \] --- ## **Step 14: Final inverse transforms** \[ x(t) = \mathcal{L}^{-1}[X(s)] = \text{Inverse of first term} + \text{Inverse of second term} \] ### First term: \[ \frac{-\frac{6}{5}(s - 1)}{(s - 1)^2 + 4} + \frac{13/10}{(s - 1)^2 + 4} \] Corresponds to: \[ -\frac{6}{5} e^{t} \cos 2t + \frac{13}{10} e^{t} \frac{\sin 2 t}{2} \] since: \[ \mathcal{L}^{-1}\left[\frac{s - a}{(s - a)^2 + b^2}\right] = e^{a t} \cos b t \] \[ \mathcal{L}^{-1}\left[\frac{b}{(s - a)^2 + b^2}\right] = e^{a t} \sin b t \] Here, \(a=1\), \(b=2\). So the first term gives: \[ -\frac{6}{5} e^{t} \cos 2t + \frac{13}{20} e^{t} \sin 2t \] ### Second term: \[ \frac{\frac{6}{5} s}{s^2 + 1} - \frac{1/10}{s^2 + 1} \] which corresponds to: \[ \frac{6}{5} \cos t - \frac{1}{10} \sin t \] --- ## **## **Final expression for \(x(t)\):** \[ \boxed{ x(t) = -\frac{6}{5} e^{t} \cos 2t + \frac{13}{20} e^{t} \sin 2t + \frac{6}{5} \cos t - \frac{1}{10} \sin t } \] --- ## **Similarly, for \(y(t)\),** you would perform inverse transforms on \(Y(s)\), which is similarly decomposed. The process would be analogous. --- ## **Summary:** **The solution for \(x(t)\) in terms of exponentials and trigonometric functions:** \[ \boxed{ x(t) = -\frac{6}{5} e^{t} \cos 2t + \frac{13}{20} e^{t} \sin 2t + \frac{6}{5} \cos t - \frac{1}{10} \sin t } \] **And similarly for \(y(t)\), once you decompose \(Y(s)\).** --- **Let me know if you'd like me to finalize \(y(t)\) explicitly!**