VIPSolutionsgive answer in 2 step with explanation at the end of each step and final answer at the end: Problem Use Laplace transforms t o solve the system x ˙ = 2 x + y + t y ˙ = 9 x + 2 y + 2 t with the initial data x ( 0 ) = 0 , y ( 0 ) = 1 .Problem Use Laplace transforms to solve the system lems §=9z+2y+2 with the initial data z (0) = 0,3 (0) = 1.
Question:
give answer in 2 step with explanation at the end of each step and final answer at the end:
Problem Use Laplace transforms
t
o
solve the system
x
˙
=
2
x
+
y
+
t
y
˙
=
9
x
+
2
y
+
2
t
with the initial data
x
(
0
)
=
0
,
y
(
0
)
=
1
.
Problem Use Laplace transforms to solve the system
lems
§=9z+2y+2
with the initial data z (0) = 0,3 (0) = 1.
This is AI generated Response.
Asked by: Pagal
Created at: 2025-06-10 09:41:37
Asked by: Pagal
Created at: 2025-06-10 09:41:37
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Let's solve the system using Laplace transforms **step by step**, with explanations at each step.
---
### **Step 1: Take Laplace Transforms of Both Equations**
Given system:
\[
\begin{cases}
\dot{x} = 2x + y + t \\
\dot{y} = 9x + 2y + 2t
\end{cases}
\]
with initial conditions \(x(0) = 0,\, y(0) = 1\).
Let \(X(s) = \mathcal{L}\{x(t)\}\), \(Y(s) = \mathcal{L}\{y(t)\}\).
**Laplace Transform of each side:**
- \(\mathcal{L}\{\dot{x}\} = sX(s) - x(0) = sX(s)\) (since \(x(0)=0\))
- \(\mathcal{L}\{2x\} = 2X(s)\)
- \(\mathcal{L}\{y\} = Y(s)\)
- \(\mathcal{L}\{t\} = \frac{1}{s^2}\)
- \(\mathcal{L}\{\dot{y}\} = sY(s) - y(0) = sY(s) - 1\)
- \(\mathcal{L}\{9x\} = 9X(s)\)
- \(\mathcal{L}\{2y\} = 2Y(s)\)
- \(\mathcal{L}\{2t\} = \frac{2}{s^2}\)
**So, the transformed system:**
\[
\begin{cases}
sX(s) = 2X(s) + Y(s) + \frac{1}{s^2} \\
sY(s) - 1 = 9X(s) + 2Y(s) + \frac{2}{s^2}
\end{cases}
\]
**Explanation:**
We converted both differential equations into algebraic equations in terms of Laplace transforms, making them easier to solve for \(X(s)\) and \(Y(s)\).
---
### **Step 2: Solve for \(X(s)\) and \(Y(s)\) in the s-domain**
Rewriting the above:
\[
\begin{cases}
sX(s) - 2X(s) - Y(s) = \frac{1}{s^2} \\
sY(s) - 2Y(s) - 9X(s) = \frac{2}{s^2} + 1
\end{cases}
\]
\[
\begin{cases}
(s-2)X(s) - Y(s) = \frac{1}{s^2} \\
-9X(s) + (s-2)Y(s) = \frac{2}{s^2} + 1
\end{cases}
\]
**Solve for \(X(s)\) and \(Y(s)\):**
From the first equation:
\[
Y(s) = (s-2)X(s) - \frac{1}{s^2}
\]
Plug into the second equation:
\[
-9X(s) + (s-2)\left[(s-2)X(s) - \frac{1}{s^2}\right] = \frac{2}{s^2} + 1
\]
\[
-9X(s) + (s-2)^2 X(s) - \frac{s-2}{s^2} = \frac{2}{s^2} + 1
\]
\[
\left[(s-2)^2 - 9\right]X(s) = \frac{2}{s^2} + 1 + \frac{s-2}{s^2}
\]
\[
\left[s^2 - 4s + 4 - 9\right]X(s) = \frac{2}{s^2} + 1 + \frac{s-2}{s^2}
\]
\[
(s^2 - 4s - 5)X(s) = 1 + \frac{2 + s - 2}{s^2}
\]
\[
(s^2 - 4s - 5)X(s) = 1 + \frac{s}{s^2}
\]
\[
(s^2 - 4s - 5)X(s) = 1 + \frac{1}{s}
\]
**Final expressions:**
\[
X(s) = \frac{1 + \frac{1}{s}}{s^2 - 4s - 5}
\]
\[
Y(s) = (s-2)X(s) - \frac{1}{s^2}
\]
**Explanation:**
We solved the algebraic system for \(X(s)\) and \(Y(s)\) in terms of \(s\). This sets up the final step: taking inverse Laplace transforms to get \(x(t)\) and \(y(t)\).
---
### **Final Answer:**
\[
\boxed{
\begin{aligned}
&X(s) = \frac{1 + \frac{1}{s}}{s^2 - 4s - 5} \\
&Y(s) = (s-2)X(s) - \frac{1}{s^2}
\end{aligned}
}
\]
You can now take the **inverse Laplace transform** to obtain \(x(t)\) and \(y(t)\). If you need the explicit time-domain solutions, let me know!
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