give answer in 2 step with explanation at the end of each step and final answer at the end:

vec y ^ prime =[ matrix -1&4\ -2&5 matrix ] vec y + [[e ^ (-4t)], [0]]

vec y(0) = [[0], [0]]

a. Fo

Question

give answer in 2 step with explanation at the end of each step and final answer at the end:

vec y ^ prime =[ matrix -1&4\ -2&5 matrix ] vec y + [[e ^ (-4t)], [0]]

vec y(0) = [[0], [0]]

a. Form the complementary solution to the homogeneous equation.

c(t) = C1

+ C2

b. Construct a particular solution by assuming the form vec y P (t)= e ^ (-4t) vec a and solving for the undetermined constant vector vec a

p(t) =

c. Form the general solution vec y (t)= vec y_{C}(t) + vec y P (t) and impose the initial condition to obtain the solution of the initial value problem.

[[y_{1}(t)], [y_{2}(t)]] =sr [1 4]- [e* > 0
= + : 0) = .
2 a EA A PEL
a. Form the complementary solution to the homogeneous equation.
=>
Ye®) =¢ | | +c | |
b. Construct a particular solution by assuming the form Vp (t) = ed and solving for the
undetermined constant vector @.
->
yet) = | |
c. Form the general solution 0) = 0) + Vp (t) and impose the initial condition to obtain
the solution of the initial value problem.
ni)
y(t)

VIPSolutions · Accepted Answer

Let's solve the system step by step.

---

## **Step 1: Complementary Solution (Homogeneous System)**

The homogeneous system is:
$$
\vec{y}' = 
\begin{bmatrix}
-1 & 4 \
-2 & 5
\end{bmatrix} 
\vec{y}
$$

**Find eigenvalues λ:**

$$
\left| \begin{matrix} -1-\lambda & 4 \ -2 & 5-\lambda \end{matrix} \right| = 0
$$
$$
(-1-\lambda)(5-\lambda) + 8 = 0
$$
$$
(-1-\lambda)(5-\lambda) + 8 = 0
\implies ( -1-\lambda )(5-\lambda) + 8 = 0
$$
$$
(-1-\lambda)(5-\lambda) + 8 = 0
\implies ( -1-\lambda )(5-\lambda ) + 8 = 0
\implies (\lambda + 1)(\lambda - 5) + 8 = 0
\implies \lambda^2 - 4\lambda + 3 = 0
$$
$$
\lambda^2 - 4\lambda + 3 = 0
\implies (\lambda - 3)(\lambda - 1) = 0
\implies \lambda = 1, 3
$$

**Find eigenvectors:**

* For **λ = 1**:
$$
\begin{bmatrix}
-2 & 4 \
-2 & 4
\end{bmatrix}
\vec{v} = 0
\implies -2v_1 + 4v_2 = 0
\implies v_1 = 2v_2
\implies \vec{v}_1 = \begin{bmatrix} 2 \ 1 \end{bmatrix}
$$

* For **λ = 3**:
$$
\begin{bmatrix}
-4 & 4 \
-2 & 2
\end{bmatrix}
\vec{v} = 0
\implies -4v_1 + 4v_2 = 0
\implies v_1 = v_2
\implies \vec{v}_2 = \begin{bmatrix} 1 \ 1 \end{bmatrix}
$$

So, the complementary solution is:
$$
\boxed{
\vec{y}_c(t) = c_1 
\begin{bmatrix}
2 \
1
\end{bmatrix}
e^{t} 
+ 
c_2 
\begin{bmatrix}
1 \
1
\end{bmatrix}
e^{3t}
}
$$

---

#### **Step 2 Explanation:**  
We found the eigenvalues and corresponding eigenvectors to write the general solution to the homogeneous system.

---

## **Step 2: Particular Solution (Nonhomogeneous System)**

Guess: $\vec{y}_p(t) = e^{-4t} \vec{a}$, with constant vector $\vec{a}$.

Plug into the equation:
$$
\frac{d}{dt}(e^{-4t}\vec{a}) = 
\begin{bmatrix}
-1 & 4 \
-2 & 5
\end{bmatrix}
e^{-4t}\vec{a} + 
\begin{bmatrix}
e^{-4t} \
0
\end{bmatrix}
$$

$$
-4e^{-4t}\vec{a} = 
\begin{bmatrix}
-1 & 4 \
-2 & 5
\end{bmatrix}
e^{-4t}\vec{a} + 
\begin{bmatrix}
e^{-4t} \
0
\end{bmatrix}
$$
Divide both sides by $e^{-4t}$:

$$
-4\vec{a} = 
\begin{bmatrix}
-1 & 4 \
-2 & 5
\end{bmatrix}
\vec{a} + 
\begin{bmatrix}
1 \
0
\end{bmatrix}
$$

$$
\left(-4I - 
\begin{bmatrix}
-1 & 4 \
-2 & 5
\end{bmatrix}
\right)
\vec{a} = 
\begin{bmatrix}
1 \
0
\end{bmatrix}
$$

$$
\left(
\begin{bmatrix}
-4 & 0 \
0 & -4
\end{bmatrix}
-
\begin{bmatrix}
-1 & 4 \
-2 & 5
\end{bmatrix}
\right)
\vec{a} = 
\begin{bmatrix}
1 \
0
\end{bmatrix}
$$

$$
\begin{bmatrix}
-3 & -4 \
2 & -9
\end{bmatrix}
\vec{a} = 
\begin{bmatrix}
1 \
0
\end{bmatrix}
$$

Let $\vec{a} = \begin{bmatrix} a_1 \ a_2 \end{bmatrix}$:

$$
-3a_1 - 4a_2 = 1 \
2a_1 - 9a_2 = 0 \implies a_1 = \frac{9}{2}a_2
$$

Plug into first equation:

$$
-3\left(\frac{9}{2}a_2\right) - 4a_2 = 1 \
-\frac{27}{2}a_2 - 4a_2 = 1 \
-\frac{27}{2}a_2 - \frac{8}{2}a_2 = 1 \
-\frac{35}{2}a_2 = 1 \
a_2 = -\frac{2}{35}
$$
$$
a_1 = \frac{9}{2}(-\frac{2}{35}) = -\frac{9}{35}
$$

So,
$$
\vec{a} = \begin{bmatrix} -\frac{9}{35} \ -\frac{2}{35} \end{bmatrix}
$$

Thus,
$$
\boxed{
\vec{y}_p(t) = e^{-4t}
\begin{bmatrix}
-\frac{9}{35} \
-\frac{2}{35}
\end{bmatrix}
}
$$

---

#### **Step 2 Explanation:**  
We assumed a particular solution, substituted it into the nonhomogeneous equation, and solved for the vector of constants.

---

## **Final Step: General Solution and Initial Condition**

Combine the solutions:
$$
\vec{y}(t) = \vec{y}_c(t) + \vec{y}_p(t)
$$

$$
\vec{y}(t) = c_1 
\begin{bmatrix}
2 \
1
\end{bmatrix}
e^{t} 
+ 
c_2 
\begin{bmatrix}
1 \
1
\end{bmatrix}
e^{3t}
+
e^{-4t}
\begin{bmatrix}
-\frac{9}{35} \
-\frac{2}{35}
\end{bmatrix}
$$

**Use initial condition $\vec{y}(0) = \begin{bmatrix} 0 \ 0 \end{bmatrix}$:**

At $t=0$:

$$
c_1
\begin{bmatrix}
2 \
1
\end{bmatrix}
+
c_2
\begin{bmatrix}
1 \
1
\end{bmatrix}
+
\begin{bmatrix}
-\frac{9}{35} \
-\frac{2}{35}
\end{bmatrix}
=
\begin{bmatrix}
0 \
0
\end{bmatrix}
$$

So,
$$
2c_1 + c_2 - \frac{9}{35} = 0 \
c_1 + c_2 - \frac{2}{35} = 0
$$

From the second equation:
$$
c_1 + c_2 = \frac{2}{35}
$$
From the first equation:
$$
2c_1 + c_2 = \frac{9}{35}
$$

Subtract the second from the first:
$$
(2c_1 + c_2) - (c_1 + c_2) = \frac{9}{35} - \frac{2}{35} \
c_1 = \frac{7}{35} = \frac{1}{5}
$$
Then,
$$
c_1 + c_2 = \frac{2}{35} \implies c_2 = \frac{2}{35} - \frac{7}{35} = -\frac{5}{35} = -\frac{1}{7}
$$

**Final Answer:**

$$
\boxed{
\begin{bmatrix}
y_1(t) \
y_2(t)
\end{bmatrix}
=
\frac{1}{5}
\begin{bmatrix}
2 \
1
\end{bmatrix}
e^{t}
-
\frac{1}{7}
\begin{bmatrix}
1 \
1
\end{bmatrix}
e^{3t}
+
e^{-4t}
\begin{bmatrix}
-\frac{9}{35} \
-\frac{2}{35}
\end{bmatrix}
}
$$

---

### **Summary of Steps and Explanations**
1. **Step 1:** Find the complementary solution using eigenvalues and eigenvectors of the matrix.
2. **Step 2:** Guess a particular solution, substitute into the equation, and solve for constants.
3. **Final:** Combine both and use the initial condition to solve for constants.

Let me know if you need this broken down into matrix form for each blank!

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