Let's solve the system using the **Laplace Transform** step by step. Given: \[ \frac{dx}{dt} = -y + t, \quad \frac{dy}{dt} = 3x + 4y - 2 - 4t, \quad x(0) = y(0) = 0 \] Let \( X(s) = \mathcal{L}\{x(t)\} \), \( Y(s) = \mathcal{L}\{y(t)\} \). ### Step 1: Take Laplace Transforms Recall: \(\mathcal{L}\left\{\frac{dx}{dt}\right\} = sX(s) - x(0)\) \(\mathcal{L}\left\{\frac{dy}{dt}\right\} = sY(s) - y(0)\) \(\mathcal{L}\{t\} = \frac{1}{s^2}\) \(\mathcal{L}\{1\} = \frac{1}{s}\) Apply Laplace to both equations: #### First Equation: \[ \mathcal{L}\left\{\frac{dx}{dt}\right\} = \mathcal{L}\{-y + t\} \] \[ sX(s) - x(0) = -Y(s) + \frac{1}{s^2} \] \(x(0) = 0\): \[ sX(s) = -Y(s) + \frac{1}{s^2} \] #### Second Equation: \[ \mathcal{L}\left\{\frac{dy}{dt}\right\} = \mathcal{L}\{3x + 4y - 2 - 4t\} \] \[ sY(s) - y(0) = 3X(s) + 4Y(s) - 2\mathcal{L}\{1\} - 4\mathcal{L}\{t\} \] \[ sY(s) = 3X(s) + 4Y(s) - \frac{2}{s} - \frac{4}{s^2} \] --- ### Step 2: Solve the System in the s-domain #### Rearranged Equations: \[ (1) \quad sX(s) + Y(s) = \frac{1}{s^2} \] \[ (2) \quad sY(s) - 4Y(s) - 3X(s) = -\frac{2}{s} - \frac{4}{s^2} \] \[ (s - 4)Y(s) - 3X(s) = -\frac{2}{s} - \frac{4}{s^2} \] --- ### Step 3: Eliminate One Variable From (1): \( Y(s) = \frac{1}{s^2} - sX(s) \) Substitute into (2): \[ (s - 4)\left[\frac{1}{s^2} - sX(s)\right] - 3X(s) = -\frac{2}{s} - \frac{4}{s^2} \] \[ \frac{s - 4}{s^2} - s(s - 4)X(s) - 3X(s) = -\frac{2}{s} - \frac{4}{s^2} \] \[ \frac{s - 4}{s^2} + \frac{4}{s^2} + \frac{2}{s} = s(s - 4)X(s) + 3X(s) \] \[ \frac{s}{s^2} + \frac{2}{s} = (s^2 - 4s + 3)X(s) \] \[ \frac{1}{s} + \frac{2}{s} = (s^2 - 4s + 3)X(s) \] \[ \frac{3}{s} = (s^2 - 4s + 3)X(s) \] --- ### Step 4: Solve for \( X(s) \) \[ X(s) = \frac{3}{s(s^2 - 4s + 3)} \] Factor denominator: \(s^2 - 4s + 3 = (s-1)(s-3)\) So, \[ X(s) = \frac{3}{s(s-1)(s-3)} \] --- ### Step 5: Partial Fraction Decomposition Set: \[ \frac{3}{s(s-1)(s-3)} = \frac{A}{s} + \frac{B}{s-1} + \frac{C}{s-3} \] Multiply both sides by \(s(s-1)(s-3)\): \[ 3 = A(s-1)(s-3) + B s(s-3) + C s(s-1) \] Let’s plug in values for \(s\): - **For \(s = 0\):** \[ 3 = A(-1)(-3) \implies 3 = 3A \implies A = 1 \] - **For \(s = 1\):** \[ 3 = B(1)(1-3) + C(1)(1-1) \implies 3 = B(1)(-2) \implies 3 = -2B \implies B = -\frac{3}{2} \] - **For \(s = 3\):** \[ 3 = C(3)(3-1) \implies 3 = C(3)(2) = 6C \implies C = \frac{1}{2} \] So, \[ X(s) = \frac{1}{s} - \frac{3/2}{s-1} + \frac{1/2}{s-3} \] --- ### Step 6: Inverse Laplace Transform to Find \( x(t) \) \[ x(t) = \mathcal{L}^{-1}\{X(s)\} = 1 - \frac{3}{2}e^{t} + \frac{1}{2}e^{3t} \] --- ### Step 7: Solve for \( Y(s) \) Recall: \[ Y(s) = \frac{1}{s^2} - sX(s) \] Calculate \(sX(s)\): \[ sX(s) = s\left(\frac{1}{s} - \frac{3/2}{s-1} + \frac{1/2}{s-3}\right) = 1 - \frac{3}{2}\frac{s}{s-1} + \frac{1}{2}\frac{s}{s-3} \] So, \[ Y(s) = \frac{1}{s^2} - \left[1 - \frac{3}{2}\frac{s}{s-1} + \frac{1}{2}\frac{s}{s-3}\right] \] \[ = \frac{1}{s^2} - 1 + \frac{3}{2}\frac{s}{s-1} - \frac{1}{2}\frac{s}{s-3} \] Let’s write \(\frac{s}{s-1}\) and \(\frac{s}{s-3}\) in a convenient form: \[ \frac{s}{s-a} = 1 + \frac{a}{s-a} \] So, \[ \frac{s}{s-1} = 1 + \frac{1}{s-1} \] \[ \frac{s}{s-3} = 1 + \frac{3}{s-3} \] Substitute back: \[ Y(s) = \frac{1}{s^2} - 1 + \frac{3}{2}\left(1 + \frac{1}{s-1}\right) - \frac{1}{2}\left(1 + \frac{3}{s-3}\right) \] \[ = \frac{1}{s^2} - 1 + \frac{3}{2} + \frac{3}{2}\frac{1}{s-1} - \frac{1}{2} - \frac{3}{2}\frac{1}{s-3} \] \[ = \frac{1}{s^2} + 1 + \frac{3}{2}\frac{1}{s-1} - \frac{3}{2}\frac{1}{s-3} \] --- ### Step 8: Inverse Laplace Transform to Find \( y(t) \) \[ y(t) = \mathcal{L}^{-1}\left\{\frac{1}{s^2}\right\} + 1 \cdot \mathcal{L}^{-1}\left\{1\right\} + \frac{3}{2} \mathcal{L}^{-1}\left\{\frac{1}{s-1}\right\} - \frac{3}{2} \mathcal{L}^{-1}\left\{\frac{1}{s-3}\right\} \] \[ = t + 1 + \frac{3}{2}e^{t} - \frac{3}{2}e^{3t} \] --- ## **Final Answer** \[ \boxed{ \begin{aligned} x(t) &= 1 - \frac{3}{2}e^{t} + \frac{1}{2}e^{3t} \\[2ex] y(t) &= t + 1 + \frac{3}{2}e^{t} - \frac{3}{2}e^{3t} \end{aligned} } \]

Certainly! Let's carefully go through each step with detailed explanations, calculations, and reasoning to solve the initial value problem using Laplace transforms and elimination method. --- ## **Problem Restatement** Given the system of differential equations: \[ \begin{cases} \displaystyle \frac{dx}{dt} = - y + t \\ \displaystyle \frac{dy}{dt} = 3x + 4y - 2 - 4t \\ \text{Initial conditions: } x(0) = 0, \quad y(0) = 0 \end{cases} \] **Goal:** Find \(x(t)\) and \(y(t)\). --- ## **Step 1: Take Laplace Transforms of Both Equations** Recall the Laplace transform of derivatives: \[ \mathcal{L}\left\{\frac{d}{dt}f(t)\right\} = sF(s) - f(0) \] Given initial conditions \(x(0) = 0\), \(y(0) = 0\). ### Transform the first equation: \[ \frac{dx}{dt} = - y + t \] Apply Laplace: \[ sX(s) - x(0) = -Y(s) + \mathcal{L}\{t\} \] Since \(x(0) = 0\): \[ sX(s) = -Y(s) + \frac{1}{s^2} \] --- ### Transform the second equation: \[ \frac{dy}{dt} = 3x + 4y - 2 - 4t \] Laplace: \[ sY(s) - y(0) = 3X(s) + 4Y(s) - 2 \cdot \frac{1}{s} - 4 \cdot \frac{1}{s^2} \] Since \(y(0) = 0\): \[ sY(s) = 3X(s) + 4Y(s) - \frac{2}{s} - \frac{4}{s^2} \] --- ## **Step 2: Write the algebraic equations** From the first: \[ sX(s) + Y(s) = \frac{1}{s^2} \quad \quad (1) \] From the second: \[ sY(s) - 4Y(s) = 3X(s) - \frac{2}{s} - \frac{4}{s^2} \] \[ (s - 4)Y(s) = 3X(s) - \frac{2}{s} - \frac{4}{s^2} \quad \quad (2) \] --- ## **Step 3: Express \(Y(s)\) from (1)** From (1): \[ Y(s) = \frac{1}{s^2} - sX(s) \] --- ## **Step 4: Substitute into (2)** Plug into (2): \[ (s - 4) \left( \frac{1}{s^2} - sX(s) \right) = 3X(s) - \frac{2}{s} - \frac{4}{s^2} \] Distribute: \[ \frac{s - 4}{s^2} - s(s - 4)X(s) = 3X(s) - \frac{2}{s} - \frac{4}{s^2} \] Express \(s(s - 4) = s^2 - 4s\): \[ \frac{s - 4}{s^2} - (s^2 - 4s) X(s) = 3X(s) - \frac{2}{s} - \frac{4}{s^2} \] Bring all \(X(s)\) terms to one side: \[ - (s^2 - 4s) X(s) - 3X(s) = - \frac{2}{s} - \frac{4}{s^2} - \frac{s - 4}{s^2} \] Factor the left: \[ - \left( s^2 - 4s + 3 \right) X(s) = - \left( \frac{2}{s} + \frac{4}{s^2} + \frac{s - 4}{s^2} \right) \] Note the right side: \[ \frac{2}{s} + \frac{4}{s^2} + \frac{s - 4}{s^2} = \frac{2}{s} + \frac{4 + s - 4}{s^2} = \frac{2}{s} + \frac{s}{s^2} = \frac{2}{s} + \frac{1}{s} \] Simplify: \[ \frac{2}{s} + \frac{1}{s} = \frac{3}{s} \] Thus: \[ - (s^2 - 4s + 3) X(s) = - \frac{3}{s} \] Multiply both sides by \(-1\): \[ (s^2 - 4s + 3) X(s) = \frac{3}{s} \] --- ## **Step 5: Solve for \(X(s)\)** \[ X(s) = \frac{3}{s (s^2 - 4s + 3)} \] Factor the quadratic: \[ s^2 - 4s + 3 = (s - 1)(s - 3) \] So: \[ X(s) = \frac{3}{s (s - 1)(s - 3)} \] --- ## **Step 6: Partial Fraction Decomposition of \(X(s)\)** Set: \[ \frac{3}{s (s - 1)(s - 3)} = \frac{A}{s} + \frac{B}{s - 1} + \frac{C}{s - 3} \] Multiply both sides by denominator: \[ 3 = A (s - 1)(s - 3) + B s (s - 3) + C s (s - 1) \] ### Find constants \(A, B, C\): - For \(s=0\): \[ 3 = A (-1)(-3) \Rightarrow 3 = 3A \Rightarrow A = 1 \] - For \(s=1\): \[ 3 = B (1) (1 - 3) \Rightarrow 3 = B \times 1 \times (-2) \Rightarrow 3 = -2 B \Rightarrow B = -\frac{3}{2} \] - For \(s=3\): \[ 3 = C (3) (3 - 1) \Rightarrow 3 = C \times 3 \times 2 \Rightarrow 3 = 6 C \Rightarrow C = \frac{1}{2} \] --- ## **Step 7: Write \(X(s)\) in terms of inverse transforms** \[ X(s) = \frac{1}{s} - \frac{3/2}{s - 1} + \frac{1/2}{s - 3} \] ### **Inverse Laplace:** \[ x(t) = \mathcal{L}^{-1}\left\{X(s)\right\} = \mathcal{L}^{-1}\left\{\frac{1}{s}\right\} - \frac{3}{2} \mathcal{L}^{-1}\left\{\frac{1}{s - 1}\right\} + \frac{1}{2} \mathcal{L}^{-1}\left\{\frac{1}{s - 3}\right\} \] From standard transforms: \[ \mathcal{L}^{-1}\left\{\frac{1}{s}\right\} = 1 \] \[ \mathcal{L}^{-1}\left\{\frac{1}{s - a}\right\} = e^{a t} \] So: \[ x(t) = 1 - \frac{3}{2} e^{t} + \frac{1}{2} e^{3t} \] --- ## **Step 8: Find \(Y(s)\)** Recall: \[ Y(s) = \frac{1}{s^2} - sX(s) \] Calculate \(sX(s)\): \[ sX(s) = s \left( \frac{1}{s} - \frac{3/2}{s - 1} + \frac{1/2}{s - 3} \right) = 1 - \frac{3}{2} \frac{s}{s - 1} + \frac{1}{2} \frac{s}{s - 3} \] Express \(\frac{s}{s - a}\): \[ \frac{s}{s - a} = 1 + \frac{a}{s - a} \] Thus: \[ \frac{s}{s - 1} = 1 + \frac{1}{s - 1} \] \[ \frac{s}{s - 3} = 1 + \frac{3}{s - 3} \] Plug back: \[ sX(s) = 1 - \frac{3}{2} \left( 1 + \frac{1}{s - 1} \right) + \frac{1}{2} \left( 1 + \frac{3}{s - 3} \right) \] Simplify: \[ sX(s) = 1 - \frac{3}{2} - \frac{3/2}{s - 1} + \frac{1}{2} + \frac{3/2}{s - 3} \] Combine constants: \[ 1 - \frac{3}{2} + \frac{1}{2} = (1 + 0.5) - 1.5 = 1.5 - 1.5 = 0 \] So: \[ sX(s) = - \frac{3/2}{s - 1} + \frac{3/2}{s - 3} \] Now, \(Y(s)\): \[ Y(s) = \frac{1}{s^2} - \left( - \frac{3/2}{s - 1} + \frac{3/2}{s - 3} \right) = \frac{1}{s^2} + \frac{3/2}{s - 1} - \frac{3/2}{s - 3} \] --- ## **Step 9: Inverse Laplace to find \(y(t)\)** \[ y(t) = \mathcal{L}^{-1}\left\{\frac{1}{s^2}\right\} + \frac{3}{2} \mathcal{L}^{-1}\left\{\frac{1}{s - 1}\right\} - \frac{3}{2} \mathcal{L}^{-1}\left\{\frac{1}{s - 3}\right\} \] Using standard transforms: \[ \mathcal{L}^{-1}\left\{\frac{1}{s^2}\right\} = t \] \[ \mathcal{L}^{-1}\left\{\frac{1}{s - a}\right\} = e^{a t} \] Result: \[ \boxed{ \boxed{ y(t) = t + \frac{3}{2} e^{t} - \frac{3}{2} e^{3t} } } \] --- ## **## **Final Solutions:** \[ \boxed{ \begin{cases} x(t) = 1 - \frac{3}{2} e^{t} + \frac{1}{2} e^{3t} \\ y(t) = t + \frac{3}{2} e^{t} - \frac{3}{2} e^{3t} \end{cases} } \] --- ## **Summary:** - We applied Laplace transforms to convert the system into algebraic equations. - Solved for \(X(s)\) using partial fractions. - Inverted the transforms to get \(x(t)\) and \(y(t)\). - The solutions satisfy initial conditions and are expressed in closed form. --- **If you have any further questions or need clarification on any step, feel free to ask!**