Question

(a) Show that $$ x=2 e^{2 t}, \quad x=e^{7 t}, $$ and $$ y=-3 e^{2 t}, \quad y=e^{7 t} $$ are solutions of the homogeneous linear system $$ \begin{aligned} &x^{\prime}=5 x+2 y \\ &y^{\prime}=3 x+4 y \end{aligned} $$ (b) Show that the two solutions defined in part (a) are linearly independent on every interval \(a \leq t \leq b\), and write the general solution of the homogeneous system of part (a). (c) Show that $$ \begin{aligned} &x=t+1 \\ &y=-5 t-2 \end{aligned} $$ is a particular solution of the nonhomogeneous linear system $$ \begin{aligned} &x^{\prime}=5 x+2 y+5 t \\ &y^{\prime}=3 x+4 y+17 t \end{aligned} $$ and write the general solution of this system.

Short answer

The general solution of the nonhomogeneous system is given by: $$ \begin{aligned} &x(t) = 2c_1e^{2t} + c_2e^{7t} + t + 1 \\ &y(t) = -3c_1e^{2t} + c_2e^{7t} - 5t - 2 \end{aligned} $$

Step-by-step solution

Verify Given Solutions for the Homogeneous System

To check if the given functions are solutions to the homogeneous linear system, substitute the functions into the system of equations and check if they satisfy the system. For \(x = 2e^{2t}\) and \(y = -3e^{2t}\): $$ x'(t) = 4e^{2t}, \quad y'(t) = -6e^{2t} $$ Substitute into the homogeneous system: $$ LHS_1 = x'(t) = 4e^{2t} $$ $$ RHS_1 = 5x(t) + 2y(t) = 5(2e^{2t}) + 2(-3e^{2t}) = 4e^{2t} $$ $$ LHS_2 = y'(t) = -6e^{2t} $$ $$ RHS_2 = 3x(t) + 4y(t) = 3(2e^{2t}) + 4(-3e^{2t}) = -6e^{2t} $$ As \(LHS_1 = RHS_1\) and \(LHS_2 = RHS_2\), the first pair of functions are solutions to the homogeneous system. For \(x = e^{7t}\) and \(y = e^{7t}\): $$ x'(t) = 7e^{7t}, \quad y'(t) = 7e^{7t} $$ Substitute into the homogeneous system: $$ LHS_1 = x'(t) = 7e^{7t} $$ $$ RHS_1 = 5x(t) + 2y(t) = 5(e^{7t}) + 2(e^{7t}) = 7e^{7t} $$ $$ LHS_2 = y'(t) = 7e^{7t} $$ $$ RHS_2 = 3x(t) + 4y(t) = 3(e^{7t}) + 4(e^{7t}) = 7e^{7t} $$ As \(LHS_1 = RHS_1\) and \(LHS_2 = RHS_2\), the second pair of functions are also solutions to the homogeneous system.

Show Linear Independence

To show that the two solutions \((2e^{2t}, -3e^{2t})\) and \((e^{7t}, e^{7t})\) are linearly independent, we will check that their Wronskian is nonzero. $$ W(t) = \begin{vmatrix} 2e^{2t} & e^{7t} \\ -3e^{2t} & e^{7t} \end{vmatrix} = 2e^{9t} - (-3e^{9t}) = 5e^{9t} \neq 0 $$ Since the Wronskian is nonzero, the two solutions are linearly independent.

General Solution of the Homogeneous System

The general solution of the homogeneous system will be a linear combination of the two linearly independent solutions. Let \(c_1, c_2\) be arbitrary constants. Then, the general solution \((x_h, y_h)\) is given by: $$ x_h(t) = c_1(2e^{2t}) + c_2(e^{7t}) = 2c_1e^{2t} + c_2e^{7t} $$ $$ y_h(t) = c_1(-3e^{2t}) + c_2(e^{7t}) = -3c_1e^{2t} + c_2e^{7t} $$

Verify the Particular Solution of the Nonhomogeneous System

To check if the given functions \(x = t + 1\) and \(y = -5t - 2\) are a particular solution of the nonhomogeneous linear system, substitute the functions into the system and check if they satisfy the system. For \(x = t + 1\) and \(y = -5t - 2\): $$ x'(t) = 1, \quad y'(t) = -5 $$ Substitute into the nonhomogeneous system: $$ LHS_1 = x'(t) = 1 $$ $$ RHS_1 = 5x(t) + 2y(t) + 5t = 5(t + 1) + 2(-5t - 2) + 5t = 1 $$ $$ LHS_2 = y'(t) = -5 $$ $$ RHS_2 = 3x(t) + 4y(t) + 17t = 3(t + 1) + 4(-5t - 2) + 17t = -5 $$ As \(LHS_1 = RHS_1\) and \(LHS_2 = RHS_2\), the given functions are a particular solution of the nonhomogeneous system.

General Solution of the Nonhomogeneous System

The general solution of the nonhomogeneous system is given by the sum of the homogeneous general solution and the particular solution found in step 4. For \(x(t)\): $$ x(t) = x_h(t) + x_p(t) = 2c_1e^{2t} + c_2e^{7t} + t + 1 $$ For \(y(t)\): $$ y(t) = y_h(t) + y_p(t) = -3c_1e^{2t} + c_2e^{7t} - 5t - 2 $$ So, the general solution of the nonhomogeneous system is: $$ \begin{aligned} &x(t) = 2c_1e^{2t} + c_2e^{7t} + t + 1 \\ &y(t) = -3c_1e^{2t} + c_2e^{7t} - 5t - 2 \end{aligned} $$

Key concepts

Homogeneous Linear System

A Homogeneous Linear System of differential equations takes on a form where the right-hand side of the equations consists of just zero. This system can be generally expressed as \[\begin{aligned}&x' = ax + by, \&y' = cx + dy,\end{aligned}\]where the solutions are sought such that they satisfy both equations simultaneously without any external input or forcing function. To verify that a given set of functions is a solution, one substitutes these functions into the left-hand side (LHS) of the equations and differentiates accordingly. The resulting expressions for the derivatives should match the right-hand side (RHS) when the original functions are substituted back into the equations. If these conditions are met, then the functions are indeed solutions to the homogeneous system, just as in the provided exercise.

Nonhomogeneous Linear System

A Nonhomogeneous Linear System differs from its homogeneous counterpart as it includes an external input or forcing function, making the right-hand side non-zero:\[\begin{aligned}&x' = ax + by + f(t), \&y' = cx + dy + g(t),\end{aligned}\]where '\(f(t)\)' and '\(g(t)\)' represent the external inputs. Similar to the homogeneous case, one tests a candidate particular solution by substitution into the LHS and then comparing with the RHS which includes the forcing functions. If these sides match up as well, then the provided functions indeed represent a particular solution to the nonhomogeneous system. The exercise clearly demonstrates this by proving that '\((t + 1, -5t - 2)\)' satisfies the given nonhomogeneous linear system when appropriate substitutions are made.

General Solution

The General Solution of a differential equation system encapsulates all possible solutions and is typically represented as a combination of particular and homogeneous solutions. For homogeneous systems, it is a linear combination of all linearly independent solutions. Each of these independent solutions is typically multiplied by an arbitrary constant, implying there is an infinite spectrum of solutions based on the choice of these constants. In nonhomogeneous cases, the general solution is the sum of the homogeneous general solution and any particular solution to the nonhomogeneous system. This effectively means that the general behavior (captured by the homogeneous solution) is adjusted by the particular solution that accounts for the additional external inputs, as seen in the final step of the problem where the general solution of the nonhomogeneous system was constructed.

Wronskian

The Wronskian is a function of time used to determine if a set of solutions is linearly independent. A non-zero Wronskian over an interval implies that the solutions are indeed independent on that interval. To compute the Wronskian, one arranges the solutions and their derivatives into a matrix and calculates its determinant. If the determinant, which is the Wronskian, does not equal zero, the functions form a linearly independent set. For the exercise in question, the Wronskian of two solutions '\(2e^{2t}\)' and '\(e^{7t}\)' and their derivatives was calculated, resulting in '\(5e^{9t}\)', which is always non-zero, thereby confirming their linear independence.
In summary, the Wronskian is a fundamental concept that plays a crucial role in understanding whether solutions to differential equations provide a full breadth of the behavior captured by the underlying system.