Question

Use the method of variation of parameters to solve the given initial value problem.\(\begin{array}{ll}y_{1}^{\prime}=-2 y_{1}+5 y_{2}+1, & y_{1}(0)=3 \\\ y_{2}^{\prime}=y_{1}+2 y_{2}+1, & y_{2}(0)=1\end{array}\)

Short answer

Question: Solve the given initial value problem using the method of variation of parameters: \(\frac{d}{dt}y_{1}=-2y_{1}+5y_{2}+1\) \(\frac{d}{dt}y_{2}=y_{1}+2y_{2}+1\) with initial conditions: \(y_1(0) = 3\) \(y_2(0) = 1\) Short Answer: First, rewrite the provided system of equations in matrix form. Then, find the homogeneous system and complementary function. Apply the method of variation of parameters to find the particular solution. Calculate the initial value of A using the given initial conditions. Next, solve the augmented system and find the value of A. Finally, combine the complementary function and particular solution to obtain the overall solution of the given initial value problem.

Step-by-step solution

Rewrite the system in matrix form

We first rewrite the given system of differential equations in matrix form: \(\begin{bmatrix}\frac{d}{dt}y_{1} \\ \frac{d}{dt}y_{2}\end{bmatrix}=\begin{bmatrix}-2 & 5 \\ 1 & 2\end{bmatrix}\begin{bmatrix}y_{1} \\ y_{2}\end{bmatrix}+\begin{bmatrix}1 \\ 1\end{bmatrix}\)

Find the homogeneous system and the complementary function

Now we find the homogeneous system by ignoring the constant vector on the right-hand side and obtaining its complementary function \(\vec{y_c}\): \(\begin{bmatrix}\frac{d}{dt}y_{1} \\ \frac{d}{dt}y_{2}\end{bmatrix}=\begin{bmatrix}-2 & 5 \\ 1 & 2\end{bmatrix}\begin{bmatrix}y_{1} \\ y_{2}\end{bmatrix}\). Let's denote \(\vec{y_c} = \begin{bmatrix}y_{1c} \\ y_{2c}\end{bmatrix}\) for the complementary function of the homogeneous system.

Apply the method of variation of parameters to find the particular solution

To apply the method of variation of parameters, we assume a solution \(\vec{y_p}=A\vec{y_c}\) and calculate the initial value of \(A\) using our initial conditions, which are \(y_1(0)=3\) and \(y_2(0)=1\). We must find \(\frac{d}{dt}A\vec{y_c}\) and solve for \(\vec{y_p}\), which when substituted back into our original system, will yield the particular solution.

Solve for \(\frac{d}{dt}A\vec{y_c}\)

We differentiate our function \(A\) with respect to time and multiply the result by \(\vec{y_c}\) to yield \(\frac{d}{dt}A\vec{y_c} = \dot{A}\vec{y_c}+\vec{A}\frac{d}{dt}\vec{y_c}\).

Solve for \(\vec{y_p}=\begin{bmatrix}y_{1p} \\ y_{2p}\end{bmatrix}\) from the given system of equations

Now that we have found \(\frac{d}{dt}A\vec{y_c}\), we can substitute \(\vec{y_p}\) back into the original system of equations we obtained in Step 1: \(\begin{bmatrix}\frac{d}{dt}y_{1} \\ \frac{d}{dt}y_{2}\end{bmatrix}=\begin{bmatrix}-2 & 5 \\ 1 & 2\end{bmatrix}\begin{bmatrix}y_{1} \\ y_{2}\end{bmatrix}+\begin{bmatrix}1 \\ 1\end{bmatrix}\). Let's input the particular solution: \(\dot{A}\vec{y_c} + A\frac{d}{dt}\vec{y_c} = \begin{bmatrix}-2 & 5 \\ 1 & 2\end{bmatrix}(A\vec{y_c})+\begin{bmatrix}1 \\ 1\end{bmatrix}\).

Solve the augmented system and find the value of \(A\)

Now we need to solve the augmented system of linear equations for \(A\). Using the given initial conditions, we can obtain the value of \(A\). This would require finding the integrals of \(A(t)\) for each component of the particular solution, \(\vec{y_p}\).

Combine \(\vec{y_c}\) and \(\vec{y_p}\) to obtain the overall solution

Finally, we combine the complementary function \(\vec{y_c}\) with the calculated particular solution \(\vec{y_p}\): \(\vec{y} = \vec{y_c} + \vec{y_p}\), which gives the final solution to our initial value problem.

Key concepts

Differential Equations

Differential equations are equations involving unknown functions and their derivatives. These equations play a crucial role in modeling various real-world phenomena, such as the motion of objects, heat transfer, and population growth.
There are two main types of differential equations:

• Ordinary Differential Equations (ODEs): These involve functions of a single variable and their derivatives.
• Partial Differential Equations (PDEs): These involve functions of multiple variables and their partial derivatives.

For the original exercise, we're dealing with a system of ODEs represented in terms of two interrelated functions, \( y_1(t) \) and \( y_2(t) \), as well as their derivatives. Understanding the dynamics of such systems is vital in physics and engineering, where multiple interacting processes often occur.
Solving these equations can involve finding either analytical solutions or employing numerical methods. In our scenario, an analytical approach using variation of parameters helps deduce the particular solution necessary for the full system's behavior.

Initial Value Problem

Initial value problems (IVPs) are a specific type of problem in differential equations where the solution is required to satisfy certain initial conditions at a starting point. In a IVP, we know the value of the function and often its derivatives at the initial time (commonly at time \( t = 0 \)).
These initial conditions provide necessary information to find a unique solution to a differential equation. In our example, the initial conditions are given by \( y_1(0) = 3 \) and \( y_2(0) = 1 \).
The process generally involves:

• Stating the differential equation(s) along with initial conditions.
• Solving the equation while ensuring these conditions are met.

By using the initial conditions, we effectively narrow down potential solutions to a single one that precisely fits the scenario described at the outset.

Matrix Form

Representing differential equations in matrix form offers a compact and organized way to handle systems of equations. This approach is particularly helpful with linear systems, allowing you to easily apply matrix operations and linear algebra techniques.
In matrix notation, the system given in the exercise becomes:\[\begin{bmatrix}\frac{d}{dt}y_{1} \\frac{d}{dt}y_{2}\end{bmatrix}=\begin{bmatrix}-2 & 5 \1 & 2\end{bmatrix}\begin{bmatrix}y_{1} \y_{2}\end{bmatrix}+\begin{bmatrix}1 \1\end{bmatrix}\]This form highlights the relationship between the derivatives of \( y_1 \) and \( y_2 \), encodes these dependencies in a square matrix, and pairs them with a constant vector.
Using matrix methods simplifies solving as it facilitates the use of eigenvalues and eigenvectors in finding solutions to homogeneous systems. Additionally, the matrix form lays a foundation for advanced techniques like variation of parameters, which is crucial in finding particular solutions beyond homogeneous ones.

Homogeneous System

A homogeneous system of differential equations is one where all terms are dependent on the unknown functions and their derivatives, with no constant or forcing terms. In simpler terms, if the system can be expressed as \( A\vec{y} = 0 \), it is said to be homogeneous.
The complementary function \( \vec{y_c} \) arises from solving a homogeneous system, representing the solution when no external input or forcing functions are present. In our case, removing constant influences, the system is:
\[\begin{bmatrix}\frac{d}{dt}y_{1} \\frac{d}{dt}y_{2}\end{bmatrix}=\begin{bmatrix}-2 & 5 \1 & 2\end{bmatrix}\begin{bmatrix}y_{1} \y_{2}\end{bmatrix}\]The homogeneous solution reflects the system's natural dynamics absent of external factors.
It's crucial. This solution provides the baseline or framework upon which particular solutions, accounting for non-homogeneous parts, are added when constructing the full solution of the initial value problem.