Question

Consider the initial value problem $$ \mathbf{y}^{\prime}=\left[\begin{array}{ll} 1 & t \\ t^{2} & 1 \end{array}\right] \mathbf{y}+\mathbf{g}(t), \quad \mathbf{y}(1)=\left[\begin{array}{r} 2 \\ -1 \end{array}\right] $$ Suppose we know that $$ \mathbf{y}(t)=\left[\begin{array}{c} t+\alpha \\ t^{2}+\beta \end{array}\right] $$ is the unique solution. Find \(\mathbf{g}(t)\) and the constants \(\alpha\) and \(\beta\).

Short answer

Question: Determine the explicit form of the vector function \(\mathbf{g}(t)\) and the values of α and β in the given initial value problem: $ \mathbf{y}^\prime(t) = \left[\begin{array}{ll} 1 & t \\ t^2 & 1 \end{array}\right] \mathbf{y}(t) + \mathbf{g}(t), \;\; \mathbf{y}(1) = \left[\begin{array}{r} 2 \\ -1 \end{array}\right], \;\; \mathbf{y}(t) = \left[\begin{array}{c} t+\alpha \\ t^2+\beta \end{array}\right]. $ Answer: After solving the initial value problem, the explicit form of the vector function \(\mathbf{g}(t)\) and the values of α and β are: - α = 1 - β = -2 - $\mathbf{g}(t) = \left[\begin{array}{c} -1 - t^3 + 2t \\ 2t - t^3 - t - t^2 + 2 \end{array}\right]$.

Step-by-step solution

Substitute the given solution

First, let's find the derivative of the given solution: $$\mathbf{y}(t) = \left[\begin{array}{c} t+\alpha \\ t^2+\beta \end{array}\right] \Rightarrow \mathbf{y}^\prime(t) = \left[\begin{array}{c} 1 \\ 2t \end{array}\right]$$ Now, substitute it into the IVP: $$ \left[\begin{array}{c} 1 \\ 2t \end{array}\right] = \left[\begin{array}{ll} 1 & t \\ t^2 & 1 \end{array}\right] \left[\begin{array}{c} t+\alpha \\ t^2+\beta \end{array}\right] + \mathbf{g}(t) $$

Equate the resulting equation and find \(\mathbf{g}(t)\)

Calculate the matrix product: $$ \left[\begin{array}{c} 1 \\ 2t \end{array}\right] = \left[\begin{array}{ll} 1 & t \\ t^2 & 1 \end{array}\right] \left[\begin{array}{c} t+\alpha \\ t^2+\beta \end{array}\right] + \mathbf{g}(t) = \left[\begin{array}{c} 1(t+\alpha) + t(t^2+\beta) \\ t^2(t+\alpha) + 1(t^2+\beta) \end{array}\right] + \mathbf{g}(t) $$ Simplifying it further gives: $$ \left[\begin{array}{c} 1 \\ 2t \end{array}\right] = \left[\begin{array}{c} t+\alpha + t^3+\beta t \\ t^3 + t\alpha + t^2 + \beta \end{array}\right] + \mathbf{g}(t) $$ Now, equate the two vectors: $$ \left[\begin{array}{c} 1 \\ 2t \end{array}\right] = \left[\begin{array}{c} 1+\alpha + t^3+\beta t \\ t^3 + t\alpha + t^2 + \beta \end{array}\right] + \mathbf{g}(t) $$ So, $$ \mathbf{g}(t) = \left[\begin{array}{c} 1 \\ 2t \end{array}\right] - \left[\begin{array}{c} 1+\alpha + t^3+\beta t \\ t^3 + t\alpha + t^2 + \beta \end{array}\right] = \left[\begin{array}{c} -\alpha - t^3 - \beta t \\ 2t - t^3 - t\alpha - t^2 - \beta \end{array}\right] $$

Substitute the initial conditions and find α and β

The given initial condition is $\mathbf{y}(1) = \left[\begin{array}{r} 2 \\ -1 \end{array}\right]$. Substitute it into the given solution: $$ \left[\begin{array}{r} 2 \\ -1 \end{array}\right] = \left[\begin{array}{c} 1+\alpha \\ 1+\beta \end{array}\right] $$ By equating the components, we get the system of equations: $$ \begin{cases} 1+\alpha = 2 \\ 1+\beta = -1 \end{cases} \Rightarrow \begin{cases} \alpha = 1 \\ \beta = -2 \end{cases} $$ Now we got the values for α and β, and the explicit form of \(\mathbf{g}(t)\). The final expressions are: - α = 1 - β = -2 - $\mathbf{g}(t) = \left[\begin{array}{c} -1 - t^3 + 2t \\ 2t - t^3 - t - t^2 + 2 \end{array}\right]$.

Key concepts

System of Differential Equations

When tackling differential equations, a system refers to a set that contains more than one equation involving derivatives of several functions. Each function is associated with a different variable. These systems can appear in many fields: physics, engineering, and economics, to name a few.

In the context of the original exercise, we dealt with a system of two first-order linear differential equations. Systems can often be written in matrix form, which simplifies solving them and allows us to use various mathematical tools.


Understanding how systems are structured and relate to each other is essential for solving complex problems in both theoretical and applied mathematics. The main advantage of breaking them down into simpler, individual parts is it makes them easier to interpret and resolve.

Matrix Differential Equations

Matrix differential equations are a powerful tool in simplifying and solving systems of differential equations. By using matrices, we can handle complex interactions between different variables and their derivatives more easily.

A matrix differential equation often looks something like \( \mathbf{y}' = A(t) \mathbf{y} + \mathbf{g}(t) \), where \( \mathbf{y} \) is a vector of functions, \( A(t) \) is a matrix with time-dependent elements, and \( \mathbf{g}(t) \) is a vector of other functions.

• Matrix A(t): Contains coefficients that dictate how each function in the system interacts with the others.
• Vector \( \mathbf{g}(t) \): Represents the non-homogeneous part, and can be thought of as external inputs affecting the system.

For the initial value problem in the exercise, converting the problem into matrix form helps in applying systematic approaches to solve it. Once in matrix form, various techniques, including eigenvalue methods or numerical solvers, can be used to find solutions.

Solution of Differential Equations

Solving differential equations typically involves finding unknown functions that satisfy the given equations. Particularly, an Initial Value Problem (often abbreviated as IVP) provides additional conditions at a specific point in time, commonly used to find a unique solution.

The solution process can be summarized as follows:

• Finding the Derivative: Start by taking the derivative of the proposed solution, as was done in the exercise with the given \( \mathbf{y}(t) \).
• Substitute Back: Plug this derivative back into the original differential equation to see how it fits into the system.
• Simplification: Simplify to find the particular form of unknowns or functions like \( \mathbf{g}(t) \) and constants like \( \alpha \) and \( \beta \).
• Initial Conditions: Use the initial values given in the problem to solve for these constants or functions completely.

This approach efficiently narrows down possibilities, leading us to a solution that both satisfies the differential equations and respects the initial conditions given.

In mathematical problems, it's crucial to remember that each step builds on the one before it, ensuring that the final answer is both accurate and meaningful.