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Find the solution of the initial value problem $$ u^{\prime \prime}+u=F(t), \quad u(0)=0, \quad u^{\prime}(0)=0 $$ where $$ F(t)=\left\\{\begin{array}{ll}{F_{0}(2 \pi-t),} & {0 \leq t \leq \pi} \\ {-0} & {(2 \pi-t),} & {\pi

Short Answer

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Question: Find the solution u(t) of the initial value problem for a second-order linear differential equation with a piecewise function F(t) given by, $$u^{\prime \prime}(t) + u(t) = F(t)$$ with, $$F(t)=\begin{cases} F_{0}(2 \pi - t), & 0 \leq t \leq \pi\\ -F_{0}(2 \pi - t), & \pi< t \leq 2 \pi\\ 0, & 2\pi < t \end{cases}$$ and initial conditions \(u(0)=0\) and \(u'(0)=0\).

Step by step solution

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01

Identify the intervals of F(t)

F(t) is defined as a piecewise function on three intervals: 1. \(0 \leq t \leq \pi\) : \(F(t)=F_{0}(2 \pi - t)\) 2. \(\pi< t \leq 2 \pi\) : \(F(t)=-F_{0}(2 \pi - t)\) 3. \(2\pi < t\): \(F(t)=0\) Now, we will solve the differential equation separately for each interval.
02

Solve the differential equation for the first interval

On the interval \(0 \leq t \leq \pi\), we have \(F(t) = F_0(2\pi - t)\). So, the differential equation becomes: $$u^{\prime \prime} + u = F_0(2\pi - t)$$ To solve this equation for \(u(t)\), we can use the method of variation of parameters. We first solve for the complementary solution \(u_c(t)\). The associated homogeneous equation is: $$u_c^{\prime \prime} + u_c = 0$$ The generic solution for the homogeneous equation is: $$u_c(t) = C_1 \sin(t) + C_2 \cos(t)$$ Then, we need to find the particular solution, \(u_p\), such that: $$u(t) = u_c(t) + u_p(t)$$ For the particular solution, we can guess: $$u_p(t) = AtB$$ From there, we can find A and B by taking the second and first derivatives of \(u_p\) and substituting them into the original differential equation. Finally, we will get the general solution for the first interval.
03

Solve the differential equation for the second interval

On the interval \(\pi< t \leq 2 \pi\), we have \(F(t) = -F_0(2\pi - t)\). So, the differential equation becomes: $$u^{\prime \prime} + u = -F_0(2\pi - t)$$ Following the same procedure as in the first interval (using variation of parameters), we can find the general solution for the second interval.
04

Solve the differential equation for the third interval

On the interval \(2\pi < t\), we have \(F(t) = 0\). So, the differential equation becomes: $$u^{\prime \prime} + u = 0$$ This is a homogeneous equation, so the general solution can be directly found as: $$u(t) = C_3 \sin(t) + C_4 \cos(t)$$
05

Combine the solutions and apply initial conditions

Now that we have the general solutions for each interval, we can combine them into a single solution by enforcing the continuity condition on \(u\) and \(u'\) at the breakpoints of the intervals (\(t=\pi\) and \(t=2\pi\)). This will give us a system of equations to solve for the constants in our general solutions. Finally, we apply the initial conditions \(u(0)=0\) and \(u'(0)=0\) to completely determine the solution of the initial value problem.

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Piecewise Functions
A piecewise function is a type of function that is defined by multiple sub-functions.
Each sub-function applies to a certain interval of the main function's domain. In the context of the problem at hand, we have the piecewise function \(F(t)\):
  • For \(0 \leq t \leq \pi\), \(F(t) = F_{0}(2\pi - t)\)
  • For \(\pi < t \leq 2\pi\), \(F(t) = -F_{0}(2\pi - t)\)
  • For \(t > 2\pi\), \(F(t) = 0\)
Each part of this piecewise function behaves differently within its defined interval. This allows the function to exhibit different behaviors, which is handy for modeling scenarios where the outcomes change in a stepwise manner with respect to the independent variable, \(t\).
In mathematical problems like this one, it is crucial to examine each piece and determine the solution separately before combining them.
Variation of Parameters
Variation of parameters is a method used to find particular solutions to non-homogeneous differential equations.
Unlike other methods, it does not require a guess for the form of the solution. Instead, it involves using known solutions to the related homogeneous equation to construct a particular solution.
When applying the variation of parameters, we use the solutions from the homogeneous equation, which in this case is \(u^{\prime \prime} + u = 0\). These solutions, \(\sin(t)\) and \(\cos(t)\), help us form the general solution:
  • The complementary solution is \(u_c(t) = C_1 \sin(t) + C_2 \cos(t)\)
  • Using variation of parameters, we find a particular solution \(u_p(t)\) for our non-homogeneous equation.
Together, these solutions give us the full solution to the differential equation in each interval. By splitting them according to each piece of \(F(t)\) and applying variation of parameters, we can construct specific solutions for different time intervals.
Homogeneous Equation
A homogeneous equation in the context of differential equations is when the equation is set to zero (i.e., \(u^{\prime \prime} + u = 0\)).
Solving this equation is crucial because its solutions form the basis for finding particular solutions to the more complex non-homogeneous variations.
The solutions of this basic form are called the complementary solutions, \(u_c(t)\), and are typically composed of terms like \(\sin(t)\) and \(\cos(t)\). These basic trigonometric functions arise from assuming exponential forms and the characteristic equation.
  • The homogeneous equation gives the foundation onto which particular solutions due to external forces, like in \(F(t)\), are added.
  • These form the basis for creating the variation of parameters technique.
Understanding the homogeneous equation is key since its solutions determine the structure of possible solutions for any non-homogeneous differential equation.
Continuity Conditions
Continuity conditions ensure a smooth transition for piecewise solutions. This means both the function \(u(t)\) and its derivative \(u'(t)\) must not exhibit abrupt changes at the breakpoints (such as \(t = \pi\) and \(t = 2\pi\)). This is crucial for a function that mimics real-world scenarios where abrupt or non-smooth solutions do not realistically represent physical phenomena.
  • Ensuring continuity means that at each breakpoint, the left-hand limit equals the right-hand limit for both \(u(t)\) and \(u'(t)\).
  • These conditions can be understood as boundary conditions specific to transitions between intervals.
  • Checking continuity imposes additional equations on the constants derived from the piecewise solutions, helping complete the overall solution.
By enforcing these continuity conditions, we ensure that the combined solution maintains a consistent, smooth behavior throughout its entire domain.

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Most popular questions from this chapter

In this problem we indicate an alternate procedure? for solving the differential equation $$ y^{\prime \prime}+b y^{\prime}+c y=\left(D^{2}+b D+c\right) y=g(t) $$ $$ \begin{array}{l}{\text { where } b \text { and } c \text { are constants, and } D \text { denotes differentiation with respect to } t \text { , Let } r_{1} \text { and } r_{2}} \\ {\text { be the zeros of the characteristic polynomial of the corresponding homogeneous equation. }} \\ {\text { These roots may be real and different, real and equal, or conjugate complex numbers. }} \\ {\text { (a) Verify that } \mathrm{Eq} \text { . (i) can be written in the factored form }}\end{array} $$ $$ \left(D-r_{1}\right)\left(D-r_{2}\right) y=g(t) $$ $$ \begin{array}{l}{\text { where } r_{1}+r_{2}=-b \text { and } r_{1} r_{2}=c} \\\ {\text { (b) Let } u=\left(D-r_{2}\right) y . \text { Then show that the solution of } \mathrm{Eq}(\mathrm{i}) \text { can be found by solving the }} \\\ {\text { following two first order equations: }}\end{array} $$ $$ \left(D-r_{1}\right) u=g(t), \quad\left(D-r_{2}\right) y=u(t) $$

Use the method outlined in Problem 28 to solve the given differential equation. $$ t^{2} y^{\prime \prime}-2 t y^{\prime}+2 y=4 t^{2}, \quad t>0 ; \quad y_{1}(t)=t $$

(a) Determine a suitable form for \(Y(t)\) if the method of undetermined coefficients is to be used. (b) Use a computer algebra system to find a particular solution of the given equation. $$ y^{\prime \prime}+3 y^{\prime}+2 y=e^{f}\left(t^{2}+1\right) \sin 2 t+3 e^{-t} \cos t+4 e^{t} $$

Verify that the given functions \(y_{1}\) and \(y_{2}\) satisfy the corresponding homogeneous equation; then find a particular solution of the given nonhomogeneous equation. In Problems 19 and \(20 g\) is an arbitrary continuous function. $$ (1-t) y^{\prime \prime}+t y^{\prime}-y=2(t-1)^{2} e^{-t}, \quad 0

Find the general solution of the given differential equation. $$ y^{\prime \prime}+5 y^{\prime}=0 $$

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