Question

In many physical problems the nonhomogencous term may be specified by different formulas in different time periods. As an example, determine the solution \(y=\phi(t)\) of $$ y^{\prime \prime}+y=\left\\{\begin{array}{ll}{t,} & {0 \leq t \leq \pi} \\\ {\pi e^{x-t},} & {t>\pi}\end{array}\right. $$ $$ \begin{array}{l}{\text { satisfying the initial conditions } y(0)=0 \text { and } y^{\prime}(0)=1 . \text { Assume that } y \text { and } y^{\prime} \text { are also }} \\ {\text { continuous at } t=\pi \text { . Plot the nonhomogencous term and the solution as functions of time. }} \\ {\text { Hint: First solve the initial value problem for } t \leq \pi \text { ; then solve for } t>\pi \text { , determining the }} \\ {\text { constants in the latter solution from the continuity conditions at } t=\pi \text { . }}\end{array} $$

Short answer

Question: Solve the given second-order nonhomogeneous differential equation with initial conditions. The equation is \(y'' + y = \) \(t, for 0 \leq t \leq \pi\) and \( y''+ y = \pi e^{x-t}, for t > \pi\), with initial conditions \(y(0) = 0\) and \(y'(0) = 1\). Answer: The solution to the given differential equation is: \(y(t) = \left\\{\begin{array}{ll}{\sin t + t,} & {0 \leq t \leq \pi} \\\ {\frac{\pi}{4}e^{x-t}\cos t + \left(1 - \frac{\pi}{4}e^{x-t}\right)\sin t,} & {t>\pi}\end{array}\right.\)

Step-by-step solution

Solve the homogeneous part of the differential equation

Take the characteristic equation from the homogeneous part of the given differential equation: \(r^2 + 1 = 0\). Solving this, we get the complex roots \(r=-i\) and \(r=i\). Therefore, the complementary function (homogeneous solution) is \(Y_c = A\cos t + B\sin t\), where A and B are constants.

Solve the particular case for \(0 \leq t \leq \pi\)

In this interval, the nonhomogeneous term is \(t\). Therefore, we need to find the particular solution \(Y_p\) to the equation: \(y'' + y = t\). We assume the particular solution to be in the form of \(Y_p = At + B\), since we have a linear term for the nonhomogeneous part. Differentiating the assumed particular solution twice: \(Y'_p = A\) \(Y''_p = 0\) Substituting this back into the differential equation: \(0 + At + B = t\) Comparing coefficients, we have: \(A = 1\) and \(B = 0\) So the particular solution is \(Y_p = t\). Now we combine the complementary function and the particular solution to get a general solution for this interval: \(y(t) = A\cos t + B\sin t + t\), for \(0 \leq t \leq \pi\)

Solve the particular case for \(t > \pi\)

In this interval, the nonhomogeneous term is \(\pi e^{x-t}\). Therefore, we need to find the particular solution \(Y_p\) to the equation: \(y'' + y = \pi e^{x-t}\). We assume the particular solution to be in the form of \(Y_p = Ce^{x-t}\), as the nonhomogeneous term involves an exponential function. Differentiating the assumed particular solution twice: \(Y'_p = C(x-2t)e^{x-t}\) \(Y''_p = C(5-4t)e^{x-t}\) Substituting this back into the differential equation: \(C(5-4t)e^{x-t} + Ce^{x-t}= \pi e^{x-t}\) Comparing coefficients, we have: \(C(5-4t) + C = \pi\) \(C = \frac{\pi}{4}\) So the particular solution is \(Y_p = \frac{\pi}{4}e^{x-t}\). Now we combine the complementary function and the particular solution to get a general solution for this interval: \(y(t) = A\cos t + B\sin t + \frac{\pi}{4}e^{x-t}\), for \(t > \pi\)

Apply the initial conditions to the first interval

We are given the initial conditions \(y(0) = 0\) and \(y'(0) = 1\). Apply these to the general solution for \(0 \leq t \leq \pi\): \(y(0) = A\cos 0 + B\sin 0 + 0 = A = 0\) \(y'(t) = -A\sin t + B\cos t + 1\) \(y'(0) = 1 = B\) Since A = 0 and B = 1, the solution for this interval becomes: \(y(t) = \sin t + t\), for \(0 \leq t \leq \pi\)

Apply continuity conditions at \(t=\pi\)

We need the solutions to be continuous at \(t=\pi\). Hence, \(y(\pi^-) = y(\pi^+)\) and \(y'(\pi^-) = y'(\pi^+)\). Apply these conditions to the solutions: \(y(\pi^-) = \sin \pi + \pi = \pi\) \(y(\pi^+) = A\cos \pi + B\sin \pi + \frac{\pi}{4}e^{x-\pi} = \frac{\pi}{4}e^{x-\pi} - A\) \(y'(\pi^-) = \cos \pi + 1 = 0\) \(y'(\pi^+) = -A\sin \pi + B\cos \pi - \frac{\pi}{4}e^{x-\pi} = 1 - B - \frac{\pi}{4}e^{x-\pi}\) Solving these equations, we have: \(A = \frac{\pi}{4}e^{x-\pi} - \pi\) \(B = 1 - \frac{\pi}{4}e^{x-\pi}\) So the final solution becomes: \(y(t) = \left\\{\begin{array}{ll}{\sin t + t,} & {0 \leq t \leq \pi} \\\ {\frac{\pi}{4}e^{x-t}\cos t + \left(1 - \frac{\pi}{4}e^{x-t}\right)\sin t,} & {t>\pi}\end{array}\right.\)

Key concepts

Nonhomogeneous Differential Equations

Nonhomogeneous differential equations are a type of differential equation that include a term which is not a function of the unknown function and its derivatives alone. These equations can be expressed in the form:\[ y'' + p(t)y' + q(t)y = g(t) \]where \( g(t) \) represents the nonhomogeneous or external forcing term.
Understanding these equations involves identifying two key components:

• The homogeneous part: This is the component similar to the left side of the equation, often involving functions of the dependent variable and its derivatives.
• The nonhomogeneous part: This is the function \( g(t) \) which introduces an external influence.

The main goal when solving such equations is to find the general solution, which is a combination of the solutions to the homogeneous equation and a particular solution to the nonhomogeneous equation.\[ y_{general} = y_{homogeneous} + y_{particular} \]
This separation helps to systematically address the complexity these equations present, which is critical when they vary over different intervals.

Initial Value Problems

An initial value problem (IVP) in differential equations specifies the value of the solution and its derivatives at a particular point. It can be formulated as follows:\[ y'' + y = f(t) \] with initial conditions such as \( y(0) = y_0 \) and \( y'(0) = y_1 \).
These conditions are necessary to uniquely determine the constant parameters from the general solution derived from the differential equation.
When you encounter an IVP:

• Calculate the general solution of the differential equation.
• Plug the initial values into the general solution.
• Solve for unknown constants to satisfy these conditions.

This process ensures that the solution adheres to the specific scenario or setup described by the problem, providing practical utility in real-world applications where conditions at a particular point are known.

Continuity of Solutions

Continuity of solutions in the context of differential equations ensures that solutions are smooth transitions between different states. This concept becomes critical when the equation is defined piecewise, as discontinuities can occur at transition points.
For the solutions to be continuous at these points, we must ensure:

• The value of the function on the left side of the transition equals the value on the right side.
• The derivative (slope) of the function on the left side matches the derivative on the right side.

In practice, this means applying continuity conditions like \( y(t^-) = y(t^+) \) and \( y'(t^-) = y'(t^+) \) at transition points such as \( t = \pi \).
These conditions are crucial to maintain the physical validity of the solution, especially in physical and engineering contexts where abrupt changes can lead to misleading or false interpretations of the system's behavior.

Complementary Function

The complementary function in a nonhomogeneous differential equation is derived from solving the associated homogeneous equation. This function represents the part of the solution influenced only by the system’s internal properties, absent the external nonhomogeneous term.
For example, given the homogeneous equation:\[ y'' + y = 0 \]To solve for the complementary function, we start with the characteristic equation derived from the differential equation:\[ r^2 + 1 = 0 \]The roots of this equation are complex: \( r = \pm i \).
The general solution or complementary function for these complex roots is:\[ Y_c = A \cos (t) + B \sin (t) \]Here, \( A \) and \( B \) are constants determined by initial or boundary conditions. The complementary function accounts for the natural dynamics of the system, forming a basis upon which particular solutions (accounting for specific external influences) are added to find the complete solution.