Problem 37 - Differential Equations and Linear Algebra

Question

Solve the given initial-value problem: $$\begin{array}{l} (D-1)(D-2)(D-3) y=6 e^{4 x}, y(0)=4,\\\ y^{\prime}(0)=10, y^{\prime \prime}(0)=30. \end{array}$$

Step-by-Step Solution

Answer

The solution to the given initial-value problem is: \(y(x) = -\frac{23}{121} e^x + \frac{17}{121} e^{2x} + \frac{80}{121} e^{3x} - \frac{3}{11} e^{4x}\)

Step 1: Solving the Homogeneous Differential Equation

For the homogeneous equation, we have: \((D - 1)(D - 2)(D - 3)y = 0\) To solve this, we can find the roots of the characteristic equation. The corresponding characteristic equation is: \(r^3 - 6r^2 + 11r - 6 = 0\) Upon factoring, we obtain the roots of 1, 2 and 3: \((r - 1)(r - 2)(r - 3) = 0\) Therefore, the general homogeneous solution is a linear combination of exponentials with the roots as coefficients: \(y_h(x) = C_1 e^x + C_2 e^{2x} + C_3 e^{3x}\)

Step 2: Finding the Particular Solution

To find the particular solution, we first determine an ansatz for the given non-homogeneous function, which is \(6e^{4x}\). Since this exponential function is not part of the homogeneous solution, we can use the same form for the ansatz with an additional constant: \(y_p(x) = A e^{4x}\) Taking the first, second and third derivatives, we have: \(y'_p(x) = 4A e^{4x}\) \(y''_p(x) = 16A e^{4x}\) \(y'''_p(x) = 64A e^{4x}\) Substitute this particular solution and its derivatives into the original non-homogeneous differential equation: \(((D - 1)(D - 2)(D - 3) y)_p = 6 e^{4 x}\) \((y'''_p - 6y''_p + 11y'_p - 6y_p) = 6 e^{4x}\) Plugging in the values obtained previously, we get: \((64A - 6(16A) + 11(4A) - 6A)e^{4x} = 6e^{4x}\) Simplifying, we get: \(-22A e^{4x} = 6 e^{4x}\) Now, we equate the corresponding coefficients, which yields: \(-22A = 6\) Therefore, we obtain \(A = -\frac{3}{11}\), so the particular solution is: \(y_p(x) = -\frac{3}{11} e^{4x}\)

Step 3: Applying the Initial Conditions

Now, we find the general solution by summing up the homogeneous and particular solutions: \(y(x) = y_h(x) + y_p(x) = C_1 e^x + C_2 e^{2x} + C_3 e^{3x} - \frac{3}{11} e^{4x}\) To find the constants, apply the given initial conditions: \(y(0) = 4\) \(y'(0) = 10\) \(y''(0) = 30\) Plug in x = 0 for each equation and find the values of \(C_1\), \(C_2\), and \(C_3\): \(y(0) = C_1 e^0 + C_2 e^0 + C_3 e^0 - \frac{3}{11} e^0=4\) \(4 = C_1 + C_2 + C_3 - \frac{3}{11}\) \(y'(0) = C_1 e^0 + 2C_2 e^0 + 3C_3 e^0 - \frac{12}{11} e^0 = 10\) \(10 = C_1 + 2C_2 + 3C_3 - \frac{12}{11}\) \(y''(0) = C_1 e^0 + 4C_2 e^0 + 9C_3 e^0 - \frac{48}{11} e^0 = 30\) \(30 = C_1 + 4C_2 + 9C_3 - \frac{48}{11}\) Solving these systems of equations, we get: \(C_1 = -\frac{23}{121}\) \(C_2 = \frac{17}{121}\) \(C_3 = \frac{80}{121}\) Finally, plug these constants back into the general solution: \(y(x) = -\frac{23}{121} e^x + \frac{17}{121} e^{2x} + \frac{80}{121} e^{3x} - \frac{3}{11} e^{4x}\) Therefore, the solution to the given initial-value problem is: \(y(x) = -\frac{23}{121} e^x + \frac{17}{121} e^{2x} + \frac{80}{121} e^{3x} - \frac{3}{11} e^{4x}\)

Key Concepts

Homogeneous Differential Equation

A homogeneous differential equation is one where the function and its derivatives are set equal to zero. This is a fundamental concept in differential equations because it represents the natural response of the system without any external forces or inputs. In our exercise, the homogeneous equation is \[(D - 1)(D - 2)(D - 3)y = 0\]where each term in the factorized form represents a differential operator acting on the function \(y\). The solution to this equation boils down to finding a general solution that satisfies this equation. This involves solving a related characteristic equation. Homogeneous differential equations often arise when modeling natural phenomena that, temporarily, don't have any external disturbances.

Particular Solution

A particular solution, unlike the homogeneous solution, is one that specifically satisfies the non-homogeneous differential equation. It focuses on responding specifically to the external inputs or forces. In our problem, we had the non-homogeneous component \[6 e^{4x}\] and needed a solution that makes the whole differential equation true, not just its homogeneous part.
The chosen form for the particular solution, since it isn't part of the homogeneous solution, is \[y_p(x) = A e^{4x}\]. This choice is strategic because it imitates the non-homogeneous term, allowing us to include the external influence \(6 e^{4x}\) in the function. The particular solution is determined by substituting this expression back into the differential equation and solving for constant \(A\).

Characteristic Equation

The characteristic equation plays a crucial role in solving linear differential equations. It's derived from the homogeneous part of the differential equation by assuming a solution of the form \(y = e^{rx}\). In our example, the characteristic equation is \[r^3 - 6r^2 + 11r - 6 = 0\].
Solving this polynomial helps find the roots, which are essential to building the general solution to the homogeneous differential equation. For our problem, factoring provided roots of \(1, 2,\) and \(3\), allowing us to form the solution \[y_h(x) = C_1 e^x + C_2 e^{2x} + C_3 e^{3x}\]. Understanding how to extract these roots is fundamental, as they determine the behavior and form of the solution.

Initial Conditions

Initial conditions are provided data points that help specify a unique solution to a differential equation, turning this system from a general solution to one specific path. Given in the form of \(y(0) = 4\), \(y'(0) = 10\), and \(y''(0) = 30\), these conditions enable us to solve for the constants \(C_1, C_2,\) and \(C_3\).
Applying initial conditions involves substituting them into the general solution equation, resulting in a system of equations. Solving these equations helps calculate the precise values of the constants, allowing for a fully determined, specific solution to the initial-value problem. This step is crucial for applications where specific outcomes or responses at the beginning of an event dictate future behavior, such as in physics and engineering.