Problem 48 - Differential Equations and Linear Algebra

Question

Solve the given initial-value problem. $$y^{\prime \prime}+2 y^{\prime}+y=2 e^{-t}, \quad y(0)=2, \quad y^{\prime}(0)=1$$.

Step-by-Step Solution

Answer

The solution to the given initial-value problem is \(y(t) = 2 e^{-t} + 5te^{-t}\).

Step 1: Find the Complementary Function

We will first solve the homogeneous version of the given ODE, \[y'' + 2y' + y = 0\] The characteristic equation for this ODE is: \[r^2 + 2r + 1 = 0\] This is a quadratic equation, and we can use the quadratic formula to solve for r: \[r_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] Here, \(a = 1\), \(b = 2\), and \(c = 1\). Plugging these values in: \[r_{1,2} = \frac{-2 \pm \sqrt{2^2 - 4(1)(1)}}{2(1)} = -1\] Since we have a repeated root, the complementary function is: \[y_c(t) = C_1 e^{-t} + C_2 te^{-t}\]

Step 2: Find a Particular Solution

Now, we need to find a particular solution to the non-homogeneous equation: \[y'' + 2y' + y = 2e^{-t}\] For that, we will use the method of undetermined coefficients. Since the right-hand side is an exponential function, we assume a particular solution of the form: \[y_p(t) = Ate^{-t}\] Now, we will find the first and second derivatives of this function: \[y'_p(t) = A(-e^{-t} + te^{-t})\] \[y''_p(t) = Ae^{-t} - 2Ate^{-t}\] Now, substitute these derivatives into the non-homogeneous equation: \[(Ae^{-t} - 2Ate^{-t}) + 2(A(-e^{-t} + te^{-t})) + Ate^{-t} = 2e^{-t}\] Simplifying the equation, we get: \[Ae^{-t} = 2e^{-t}\] \[A = 2\] Thus, the particular solution is: \[y_p(t) = 2te^{-t}\]

Step 3: Determine the General Solution

The general solution of the given ODE is the sum of the complementary function and the particular solution: \[y(t) = y_c(t) + y_p(t) = C_1 e^{-t} + C_2 te^{-t} + 2te^{-t}\]

Step 4: Apply the Initial Conditions

Now, we use the given initial conditions, \(y(0) = 2\) and \(y'(0) = 1\), to obtain the particular solution. From the general solution, we have: \[y(0) = C_1 e^{0} + C_2 (0) e^{0} + 2(0) e^{0}\] \[2 = C_1\] Now, differentiate the general solution with respect to t: \[y'(t) = -C_1 e^{-t} + C_2(e^{-t} - te^{-t}) - 2e^{-t} + 2te^{-t}\] Using the second initial condition, we have: \[y'(0) = -C_1 e^{0} + C_2 e^{0} - 2 e^{0} + 2(0)e^{0}\] \[1 = -2 + C_2\] \[C_2 = 3\] Now, plug in the values of \(C_1\) and \(C_2\) into the general solution: \[y(t) = 2 e^{-t} + 3te^{-t} + 2te^{-t}\] Simplify the final solution: \[y(t) = 2 e^{-t} + 5te^{-t}\] The solution to the initial-value problem is: \[y(t) = 2 e^{-t} + 5te^{-t}\]

Key Concepts

Homogeneous Differential Equation

When dealing with differential equations, the term "homogeneous" refers to an equation in which all the terms are associated with the function or its derivatives. In essence, there are no free-standing terms unrelated to the function. A typical example is:

• \(y'' + 2y' + y = 0\)

To solve a homogeneous differential equation, the goal is to find solutions where the given equation equals zero. The characteristic equation is used, often a quadratic or higher-degree polynomial. Solving this equation provides the roots, which in turn determine the form of the complementary function.
Repeated roots, as seen in our example, have special considerations, leading to terms like \(C_1 e^{-t} + C_2 te^{-t}\).

Method of Undetermined Coefficients

The Method of Undetermined Coefficients is a systematic approach to find a particular solution to a non-homogeneous differential equation. This method works well when the non-homogeneous term is of a certain kind, such as an exponential, polynomial, or sinusoidal function.
The method involves assuming a form for the particular solution based on the form of the non-homogeneous term, which in our case leads to:

• Assumption: \(y_p(t) = Ate^{-t}\)

Next, compute the necessary derivatives and substitute them back into the original differential equation. Finally, adjust the coefficients such that the left-hand side matches the non-homogeneous term on the right-hand side. Here, solving equated the coefficients of \(e^{-t}\) yielding \(A = 2\).
This results in the particular solution of \(2te^{-t}\).

Complementary Function

The complementary function, also known as the solution to the homogeneous equation, forms a core part of the general solution for linear differential equations. This function is derived by solving the associated homogeneous differential equation, where the equation is set to zero.

• For \(y'' + 2y' + y = 0\)

The characteristic equation is constructed and solved for its roots, often involving real and repeated roots. These roots dictate the form of the complementary function:

• Repeated root \(r = -1\) leads to the form: \(y_c(t) = C_1 e^{-t} + C_2 te^{-t}\)

This function represents the behavior of the system described by the differential equation when external influences are absent.

Particular Solution

A particular solution caters specifically to the non-homogeneous part of the differential equation and satisfies the equation when substituted back into it. In the example given, the non-homogeneous term is \(2e^{-t}\).

• Proposed form for the particular solution: \(y_p(t) = Ate^{-t}\)

Here, by differentiating \(y_p(t)\) and substituting these into the non-homogeneous differential equation, we determine \(A = 2\) through algebraic manipulation.
Thus, one arrives at the particular solution:\(y_p(t) = 2te^{-t}\). Together with the complementary function, it helps form the complete response of the system.