Solve $x'=Ax$ Initial Value Problem: A Simple Guide

Hey there, math enthusiasts and problem-solvers! Ever stared at a matrix differential equation, like $x' = Ax$, with an initial condition like $x(0)=\binom{6}{-3}$, and thought, "Whoa, what's going on here?" Well, you're in luck because today, we're going to demystify this beast and walk through solving one of these fascinating initial value problems (IVPs) step-by-step. These aren't just abstract mathematical exercises, guys; they're super important for modeling everything from population dynamics to the behavior of electrical circuits and even quantum mechanics. Understanding how to find the solution to an initial value problem like $x' = \begin{pmatrix}-4 & 5 \\ -5 & -4\end{pmatrix} x$ with a starting point of $x(0)=\binom{6}{-3}$ is a powerful skill. So, grab your favorite beverage, get comfy, and let's dive into the awesome world of differential equations! We're going to break down this complex problem into manageable, easy-to-understand chunks, ensuring you not only get the answer but also grasp the why behind each step. Our goal is to make sure you walk away feeling confident and ready to tackle similar challenges. We'll explore the crucial role of eigenvalues and eigenvectors, learn how to handle complex numbers like a pro, and finally, piece together a specific solution that fits our given initial conditions. This isn't just about crunching numbers; it's about building a solid conceptual foundation for understanding dynamic systems, making it incredibly valuable for anyone studying engineering, physics, economics, or pure mathematics. So, let's roll up our sleeves and get started on this exciting journey to master matrix initial value problems!

Understanding the Beast: What's an $x'=Ax$ Initial Value Problem?

Alright, first things first, let's get a handle on what we're actually solving here. When we talk about an initial value problem (IVP) for a system of differential equations, we're basically looking for a function $x(t)$ that satisfies two conditions: a differential equation and an initial condition. In our specific case, we have $x' = Ax$, where $x$ is a vector function of $t$ (time), and $A$ is a constant matrix. Our particular problem is $x' = \begin{pmatrix}-4 & 5 \\ -5 & -4\end{pmatrix} x$. This means we have a system of two coupled first-order linear differential equations. The $x'$ on the left side represents the derivative of the vector $x$ with respect to $t$. The $A$ on the right side is our coefficient matrix, which dictates how the components of $x$ change over time relative to each other. Think of $x(t)$ as tracking multiple interdependent quantities evolving over time. The second crucial piece of information is the initial condition: $x(0)=\binom{6}{-3}$. This tells us the exact state of our system at time $t=0$. Without this, we'd only find a general solution with arbitrary constants, but with it, we can pinpoint the unique specific solution that starts from that precise point. This is like knowing the starting position and velocity of a projectile – it allows you to predict its entire trajectory. These types of problems are fundamental in modeling real-world phenomena, offering predictive power in various scientific and engineering disciplines. For instance, if $x$ represented the concentrations of two chemicals reacting, the matrix $A$ would describe their reaction rates, and $x(0)$ would be their initial concentrations. Our job is to find the functions that describe how those concentrations change over time. The elegance of solving these lies in transforming a system of differential equations into an algebraic problem involving eigenvalues and eigenvectors, which are special values and vectors associated with our matrix $A$. These eigenvalues and eigenvectors are the key to unlocking the structure of the solutions, telling us about the fundamental modes of behavior of the system. They reveal whether the system grows, decays, oscillates, or spirals. So, understanding the structure of this problem is the first and most vital step in finding its solution. This isn't just a math puzzle; it's a way to describe and predict complex behaviors in the world around us. So let's get ready to decode the behavior of this system and unveil its full-time evolution! Knowing that we're dealing with a matrix $A$ that could lead to complex eigenvalues also prepares us for the oscillatory or spiral behavior often seen in dynamic systems, which is pretty cool if you ask me.

Step 1: Unlocking the Matrix's Secrets – Eigenvalues and Eigenvectors

Alright, guys, this is where the real magic begins! The first and most critical step in solving our matrix initial value problem $x' = Ax$ is to delve into the heart of our matrix $A$ and extract its eigenvalues and eigenvectors. These special numbers and vectors are the building blocks of our solution. They tell us about the fundamental directions and rates of change within our system. Our matrix is $A = \begin{pmatrix}-4 & 5 \\ -5 & -4\end{pmatrix}$. To find the eigenvalues, we need to solve the characteristic equation, which is given by $\det(A - \lambda I) = 0$, where $I$ is the identity matrix and $\lambda$ represents our mysterious eigenvalues. Let's substitute our matrix $A$ into this equation:

$A - \lambda I = \begin{pmatrix}-4 & 5 \\ -5 & -4\end{pmatrix} - \lambda \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix} = \begin{pmatrix}-4-\lambda & 5 \\ -5 & -4-\lambda\end{pmatrix}$

Now, we calculate the determinant of this new matrix:

$\det(A - \lambda I) = (-4-\lambda)(-4-\lambda) - (5)(-5) = (\lambda+4)^2 + 25 = 0$

Solving for $\lambda$:

$(\lambda+4)^2 = -25$

Taking the square root of both sides, remember we'll get two values, a positive and a negative:

$\lambda+4 = \pm\sqrt{-25} = \pm 5i$

Finally, isolate $\lambda$:

$\lambda = -4 \pm 5i$

Whoa, we've got complex eigenvalues! Don't panic, this is totally normal and actually pretty common in systems that exhibit oscillatory or spiral behavior, which is exactly what our problem's solution will likely show. Our eigenvalues are $\lambda_1 = -4 + 5i$ and $\lambda_2 = -4 - 5i$. Notice they are complex conjugates, which is always the case for real matrices. Now that we have our eigenvalues, we need to find the corresponding eigenvectors. An eigenvector $v$ associated with an eigenvalue $\lambda$ satisfies the equation $(A - \lambda I)v = 0$. We only need to find the eigenvector for one of the complex eigenvalues (say, $\lambda_1 = -4 + 5i$), because the eigenvector for its conjugate will also be its conjugate. Let's take $\lambda_1 = -4 + 5i$:

$A - \lambda_1 I = \begin{pmatrix}-4 - (-4+5i) & 5 \\ -5 & -4 - (-4+5i)\end{pmatrix} = \begin{pmatrix}-5i & 5 \\ -5 & -5i\end{pmatrix}$

Now, we solve $(A - \lambda_1 I)v = 0$ for $v = \begin{pmatrix}v_1 \\ v_2\end{pmatrix}$:

$\begin{pmatrix}-5i & 5 \\ -5 & -5i\end{pmatrix} \begin{pmatrix}v_1 \\ v_2\end{pmatrix} = \begin{pmatrix}0 \\ 0\end{pmatrix}$

From the first row, we get: $-5iv_1 + 5v_2 = 0$. Dividing by 5 gives $-iv_1 + v_2 = 0$, so $v_2 = iv_1$. We can choose a simple value for $v_1$ to find a corresponding $v_2$. Let's pick $v_1 = 1$. Then $v_2 = i$. So, our eigenvector $v_1$ corresponding to $\lambda_1 = -4 + 5i$ is:

$v_1 = \begin{pmatrix}1 \\ i\end{pmatrix}$

We've successfully extracted the fundamental characteristics of our matrix! These complex eigenvalues and eigenvectors are the cornerstone of our solution. They tell us that our system isn't just decaying or growing; it's also rotating or oscillating as it evolves. This is a crucial insight that sets apart systems with complex eigenvalues. Understanding this step deeply is paramount because it directly informs the shape and behavior of our final solution. Without correctly identifying these values, the rest of the solution won't make sense. So, give yourselves a pat on the back for navigating the complexities of complex numbers in this vital step! This groundwork ensures we can build a robust and accurate model for our dynamic system moving forward.

Step 2: Crafting the General Solution from Complex Components

Alright, team, we've got our eigenvalues and eigenvectors, which is awesome! Now, it's time to use these building blocks to construct the general solution for our system $x' = Ax$. When you have complex conjugate eigenvalues, say $\lambda = \alpha \pm i\beta$, and a corresponding complex eigenvector $v = a + ib$ (where $a$ is the real part and $b$ is the imaginary part of the eigenvector), the general real-valued solution takes a very specific and elegant form. This is super important because even though we derived complex eigenvalues and eigenvectors, our original problem expects a real-valued solution, as $x(t)$ represents real physical quantities. The magic formula for combining these complex pieces into a real solution is:

$x(t) = C_1 e^{\alpha t} (\text{Re}(v) \cos(\beta t) - \text{Im}(v) \sin(\beta t)) + C_2 e^{\alpha t} (\text{Re}(v) \sin(\beta t) + \text{Im}(v) \cos(\beta t))$

Let's break down our components:

• From $\lambda_1 = -4 + 5i$, we have $\alpha = -4$ and $\beta = 5$.
• From $v_1 = \begin{pmatrix}1 \\ i\end{pmatrix}$, we can split it into its real and imaginary parts:
  • $\text{Re}(v) = \begin{pmatrix}1 \\ 0\end{pmatrix}$
  • $\text{Im}(v) = \begin{pmatrix}0 \\ 1\end{pmatrix}$

Now, let's substitute these values into the general solution formula. Pay close attention to the order of $\cos$ and $\sin$ and the signs – it's crucial for getting this right!

$x(t) = C_1 e^{-4t} \left( \begin{pmatrix}1 \\ 0\end{pmatrix} \cos(5t) - \begin{pmatrix}0 \\ 1\end{pmatrix} \sin(5t) \right) + C_2 e^{-4t} \left( \begin{pmatrix}1 \\ 0\end{pmatrix} \sin(5t) + \begin{pmatrix}0 \\ 1\end{pmatrix} \cos(5t) \right)$

Let's simplify the vector expressions inside the parentheses:

$\begin{pmatrix}1 \\ 0\end{pmatrix} \cos(5t) - \begin{pmatrix}0 \\ 1\end{pmatrix} \sin(5t) = \begin{pmatrix}\cos(5t) \\ 0\end{pmatrix} - \begin{pmatrix}0 \\ \sin(5t)\end{pmatrix} = \begin{pmatrix}\cos(5t) \\ -\sin(5t)\end{pmatrix}$

And for the second part:

$\begin{pmatrix}1 \\ 0\end{pmatrix} \sin(5t) + \begin{pmatrix}0 \\ 1\end{pmatrix} \cos(5t) = \begin{pmatrix}\sin(5t) \\ 0\end{pmatrix} + \begin{pmatrix}0 \\ \cos(5t)\end{pmatrix} = \begin{pmatrix}\sin(5t) \\ \cos(5t)\end{pmatrix}$

Plugging these back into our general solution, we get:

$x(t) = C_1 e^{-4t} \begin{pmatrix}\cos(5t) \\ -\sin(5t)\end{pmatrix} + C_2 e^{-4t} \begin{pmatrix}\sin(5t) \\ \cos(5t)\end{pmatrix}$

This, my friends, is our general solution. It's a combination of two linearly independent solutions, each decaying exponentially due to the $e^{-4t}$ term (because $\alpha = -4$ is negative), and oscillating due to the $\cos(5t)$ and $\sin(5t)$ terms (because $\beta = 5$ is non-zero). The constants $C_1$ and $C_2$ are still unknown; they represent the flexibility in our solution. They'll be determined by our specific starting point – the initial condition. This step is a beautiful demonstration of how complex numbers, often seen as abstract, play a concrete role in describing physical phenomena like damping oscillations or spiral trajectories. The $e^{\alpha t}$ term controls the growth or decay, while the $\cos(\beta t)$ and $\sin(\beta t)$ terms dictate the oscillatory frequency. It's a powerful framework, and understanding how these components interact is key to truly grasping the dynamics of the system. We're almost there! We've transformed a differential equation problem into an elegant solution that captures the system's inherent oscillatory and decaying behavior. Now, we just need to tailor this general form to our specific initial condition, which is the final puzzle piece to reveal the unique trajectory of our system.

Step 3: Pinpointing the Specific Path – Applying the Initial Condition

Okay, we've got the general solution, which tells us all the possible ways our system could evolve. But our problem isn't just about possibilities; it's about a specific trajectory that starts from a particular point. This is where our initial condition $x(0)=\binom{6}{-3}$ comes into play! We need to use this information to find the exact values for our constants $C_1$ and $C_2$. Remember, these constants are like the