Math Problem Statement
Using the Laplace transform and showing the details of your work, solve the IVP: y1' = -y1 + 4y2, y1(0) = 3; y2' = 3y1 - 2y2, y2(0) = 4
Solution
The problem is to solve a system of linear differential equations using the Laplace transform:
y_1' = -y_1 + 4y_2, \quad y_1(0) = 3 \\ y_2' = 3y_1 - 2y_2, \quad y_2(0) = 4 \end{cases}$$ ### Step 1: Apply the Laplace Transform Let's denote $$\mathcal{L}\{y_1(t)\} = Y_1(s)$$ and $$\mathcal{L}\{y_2(t)\} = Y_2(s)$$. Recall that for any function $$y(t)$$: $$\mathcal{L}\{y'(t)\} = sY(s) - y(0).$$ Applying the Laplace transform to both differential equations: 1. For $$y_1' = -y_1 + 4y_2$$: $$sY_1(s) - y_1(0) = -Y_1(s) + 4Y_2(s).$$ Substitute $$y_1(0) = 3$$: $$sY_1(s) - 3 = -Y_1(s) + 4Y_2(s).$$ Simplifying: $$(s + 1)Y_1(s) - 4Y_2(s) = 3. \quad \text{(Equation 1)}$$ 2. For $$y_2' = 3y_1 - 2y_2$$: $$sY_2(s) - y_2(0) = 3Y_1(s) - 2Y_2(s).$$ Substitute $$y_2(0) = 4$$: $$sY_2(s) - 4 = 3Y_1(s) - 2Y_2(s).$$ Simplifying: $$-3Y_1(s) + (s + 2)Y_2(s) = 4. \quad \text{(Equation 2)}$$ ### Step 2: Solve the System of Equations Let's solve these simultaneous equations to find $$Y_1(s)$$ and $$Y_2(s)$$. 1. Multiply Equation 1 by $$s + 2$$: $$(s + 2)(s + 1)Y_1(s) - 4(s + 2)Y_2(s) = 3(s + 2).$$ 2. Multiply Equation 2 by 4: $$-12Y_1(s) + 4(s + 2)Y_2(s) = 16.$$ 3. Adding the two equations: $$(s + 2)(s + 1)Y_1(s) - 12Y_1(s) = 3(s + 2) + 16.$$ Simplify: $$(s^2 + 3s + 2 - 12)Y_1(s) = 3s + 6 + 16.$$ $$(s^2 + 3s - 10)Y_1(s) = 3s + 22.$$ $$Y_1(s) = \frac{3s + 22}{s^2 + 3s - 10}.$$ ### Step 3: Finding $$Y_2(s)$$ Now substitute $$Y_1(s) = \frac{3s + 22}{s^2 + 3s - 10}$$ into Equation 1: $$(s + 1)Y_1(s) - 4Y_2(s) = 3.$$ Solve for $$Y_2(s)$$: $$Y_2(s) = \frac{(s + 1) \frac{3s + 22}{s^2 + 3s - 10} - 3}{4}.$$ ### Step 4: Inverse Laplace Transform Find $$y_1(t)$$ and $$y_2(t)$$ by taking the inverse Laplace transform of $$Y_1(s)$$ and $$Y_2(s)$$. I will now proceed to solve these steps in detail for a full solution.The solutions for $$Y_1(s)$$ and $$Y_2(s)$$ in the Laplace domain are: $$Y_1(s) = \frac{3s + 22}{s^2 + 3s - 10}, \quad Y_2(s) = \frac{4s + 13}{s^2 + 3s - 10}.$$ ### Step 4: Inverse Laplace Transform Now weAsk a new question for Free
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Math Problem Analysis
Mathematical Concepts
Laplace Transform
Linear Differential Equations
Systems of Equations
Formulas
Laplace Transform: L{y'(t)} = sY(s) - y(0)
Inverse Laplace Transform
Theorems
Laplace Transform Theorem
Superposition Principle
Suitable Grade Level
Undergraduate Mathematics
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