Question

Consider the linear system $$ \begin{aligned} &x^{\prime}=3 x+4 y \\ &y^{\prime}=2 x+y \end{aligned} $$ (a) Show that $$ x=2 e^{5 t}, \quad x=e^{-t} $$ and $$ y=e^{5 t}, \quad y=-e^{-t} $$ are solutions of this system. (b) Show that the two solutions of part (a) are linearly independent on every interval \(a \leq t \leq b\), and write the general solution of the system. (c) Find the solution $$ \begin{aligned} &x=f(t) \\ &y=g(t) \end{aligned} $$ of the system which is such that \(f(0)=1\) and \(g(0)=2 .\) Why is this solution unique? Over what interval is it defined?

Short answer

The general solution to the given linear system is: \[x(t) = c_1(2e^{5t}) + c_2(e^{-t})\] \[y(t) = c_1(e^{5t}) + c_2(-e^{-t})\] The specific solution satisfying the initial conditions \(x(0)=1\) and \(y(0)=2\) is: \[x(t) = 2e^{5t} + e^{-t}\] \[y(t) = e^{5t} - e^{-t}\] This specific solution is unique and is defined for all \(t \in (-\infty, \infty)\).

Step-by-step solution

Verify the solutions for the system

To check whether the given solutions satisfy the system of differential equations, we need to plug them into the system and go through the following steps: a) Find the derivatives of the given solutions. b) Substitute the solutions and their derivatives into the system and verify the equations.

Derivatives of the solutions

First, find the derivatives of the given solutions, \(x(t)\) and \(y(t)\): For the first set of solutions: \(x(t)=2e^{5t} \Rightarrow x'(t)=10e^{5t}\) \(y(t)=e^{5t} \Rightarrow y'(t)=5e^{5t}\) For the second set of solutions: \(x(t)=e^{-t} \Rightarrow x'(t)=-e^{-t}\) \(y(t)=-e^{-t} \Rightarrow y'(t)=e^{-t}\)

Verify the given solutions

Substitute the solutions and their derivatives into the original system of equations: For the first set of solutions (\(x=2e^{5t}, y=e^{5t}\)): \(x' = 10 e^{5t} \equiv 3(2e^{5t}) + 4(e^{5t})\) \(y' = 5 e^{5t} \equiv 2(2e^{5t}) + (e^{5t})\) For the second set of solutions (\(x=e^{-t}, y=-e^{-t}\)): \(x' = -e^{-t} \equiv 3(e^{-t}) + 4(-e^{-t})\) \(y' = e^{-t} \equiv 2(e^{-t}) + (-e^{-t})\) Since the given solutions and their derivatives satisfy the system of differential equations, they are indeed solutions of the linear system.

Prove the linear independence

We can show that the solutions are linearly independent by analyzing their Wronskian: \[W(t) = \begin{vmatrix} 2e^{5t} & e^{-t} \\ e^{5t} & -e^{-t} \end{vmatrix} \] Calculate the determinant of the Wronskian matrix: \[W(t) = -2e^{5t}(-e^{-t}) - e^{-t}(e^{5t})\] \[W(t) = 2e^{4t} - e^{4t} = e^{4t} \neq 0\] Since the Wronskian is not equal to zero, the two solutions are linearly independent.

General solution of the system

Now that we have verified the given solutions and proved their linear independence, we can write the general solution of the system: \[x(t) = c_1(2e^{5t}) + c_2(e^{-t})\] \[y(t) = c_1(e^{5t}) + c_2(-e^{-t})\]

Solve for specific solution

To find the specific solution satisfying the initial conditions \(x(0)=1\) and \(y(0)=2\), plug in these values into the general solution: \[ x(0)=1 = c_1(2)+(c_2)(1) \] \[ y(0)=2 = c_1(1)+(c_2)(-1) \] Solving this system of linear equations for \(c_1\) and \(c_2\), we get: \[ c_1 = 1 \] \[ c_2 = 1 \] So, the specific solution is: \[ x(t) = 2e^{5t} + e^{-t} \] \[ y(t) = e^{5t} - e^{-t} \] The specific solution satisfying this initial condition is unique because the given linear system is homogeneous, and the solutions are linearly independent. The solution is defined for all \(t \in (-\infty, \infty)\), as it has no limitations on the time interval.

Key concepts

Linear Independence

In the context of systems of linear differential equations, the concept of linear independence is crucial. Two functions are considered linearly independent if there is no scalar, nontrivial combination of these functions that equals zero, except for the trivial case where all scalars are zero.

Verifying linear independence within a system of differential equations is typically done using the Wronskian determinant, a special determinant constructed from the functions and their derivatives. If the Wronskian is never zero over an interval, the functions in question are linearly independent over that interval.

Such independence is important because it ensures that the solutions to the differential equations form a basis set for the solution space, allowing us to construct the most general solution to the system.

Wronskian Determinant

The Wronskian determinant, often simply called the Wronskian, is a method used to determine whether two or more functions are linearly independent. It is defined for two functions, say f(t) and g(t), as:

\[ W(t) = \begin{vmatrix} f(t) & g(t) \ f'(t) & g'(t) \end{vmatrix} \]
For the system in our exercise, the Wronskian of the solutions was computed and found to be nonzero for all t, confirming that the solutions are indeed linearly independent. This property is fundamental when we must ensure that the general solution to the system encompasses all possible solutions.

General Solution of Differential Equations

The general solution of a system of linear differential equations represents the full set of possible solutions, including all specific instances of the equation. It is derived by combining all linearly independent solutions of the homogeneous equation, each multiplied by an arbitrary constant. For the system under consideration, the general solution includes the linear combination of the two independent solutions:

\[ x(t) = c_1(2e^{5t}) + c_2(e^{-t}) \]
\[ y(t) = c_1(e^{5t}) + c_2(-e^{-t}) \]
The constants c_1 and c_2 are determined based on initial conditions or other constraints that the solution must satisfy.

Specific Solution with Initial Conditions

A specific solution to a differential equation is one that satisfies not only the equation itself but also certain initial conditions—specific values of the function and potentially its derivatives at a given point. To find such a solution from the general solution, we substitute the initial conditions into the equation and solve for the arbitrary constants included.

For instance, with initial conditions \( x(0) = 1 \) and \( y(0) = 2 \), we determined the constants \( c_1 \) and \( c_2 \) to craft a unique specific solution that precisely fits these conditions. The uniqueness of this solution is a direct result of the functions' linear independence and the homogeneity of the system.

Homogeneous Differential Equations

A homogeneous differential equation is one where all terms are multiples of the function or its derivatives, and there is no standalone term (no constant term). In simpler terms, if all instances of the dependent variable(s) and their derivatives were removed, the equation would equal zero.

When considering the entire set of solutions to a homogeneous system, any linear combination of solutions is also a solution. This property is essential in constructing the general solution, where linear independence guarantees no redundancy in the combination of solutions. For the given system, its homogeneity allows us to confidently say that the specific solution we find is unique and valid over all real numbers, as implied by the absence of restrictions on the time interval.