Chapter 2: Problem 29
Find the general solution of the differential equation \(\frac{d y}{d t}+2 t y=\delta(t-a)\).
Short Answer
Expert verified
The general solution is \(y(t) = Ce^{-t^2} + \delta(t-a)\).
Step by step solution
01
Understand the Differential Equation
The differential equation given is \(\frac{d y}{dt} + 2ty = \delta(t-a)\). Here, \(\delta(t-a)\) is the Dirac delta function, which is zero everywhere except at \(t = a\), where it is defined to integrate to 1. This implies the right side represents an impulse at \(t = a\).
02
Homogeneous Solution
Solve \(\frac{d y_h}{dt} + 2ty_h = 0\). This is a first-order linear homogeneous differential equation. Its solution is found using separation of variables or an integrating factor.Using an integrating factor:\[ m(t) = e^{\int 2t dt} = e^{t^2} \]The homogeneous solution is thus:\[ y_h(t) = C e^{-t^2} \] where \(C\) is a constant.
03
Particular Solution with Impulse
Consider the impulse at \(t = a\). At \(t = a\), we need a solution that accounts for the impulse: \(y_p(t) = k\delta(t-a)\) which has the form of the solution under the presence of \(\delta(t-a)\). Integrate over a small interval around \(t = a\) to determine \(k\).Integrating both sides from \(a-\epsilon\) to \(a+\epsilon\) and letting \(\epsilon\to0\):\[ [y(a+\epsilon) - y(a-\epsilon)] + \int_{a-\epsilon}^{a+\epsilon} 2ty(t) dt = 1 \]Since the integral over \(2ty(t)\) from \(a-\epsilon\) to \(a+\epsilon\) vanishes as \(\epsilon\to0\), it tells that:\[ y(a+\epsilon) = y(a-\epsilon) + 1 \] Therefore, \(y_p(t) = \delta(t-a)\).
04
General Solution Construction
Combine the homogeneous and particular solutions to find the general solution:\[ y(t) = Ce^{-t^2} + \delta(t-a) \] This accounts for the initial condition discontinuity at \(t=a\). The constant \(C\) will be determined by any initial conditions provided apart from the impulse condition.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Dirac delta function
The Dirac delta function, often represented as \( \delta(t-a) \), is a unique mathematical construct. It is zero everywhere except at the point \( t = a \), where it becomes infinitely large in such a way that its integral over the entire real line is equal to 1. This property makes it ideal for representing an instantaneous impulse or spike at a specific point in time. In practical terms, imagine a function that is zero during its entire domain, except at one single point where all its strength is concentrated. Its most remarkable feature is how it integrates:
- \( \int_{-\infty}^{\infty} \delta(t-a) \ dt = 1 \)
- \( \delta(t-a)f(t) = f(a)\delta(t-a) \)
homogeneous solution
A homogeneous solution of a differential equation is a solution that satisfies the equation when the non-homogeneous (forcing) term is set to zero. For the given equation \(\frac{d y_h}{dt} + 2ty_h = 0\), the homogeneous part of the problem ignores the impulse term \( \delta(t-a) \). Here, we solve this as a first-order linear continuous differential equation. Using an integrating factor provides a systematic method:
- The integrating factor is \( m(t) = e^{\int 2t \, dt} = e^{t^2} \)
- The homogeneous solution becomes \( y_h(t) = Ce^{-t^2} \) where \( C \) is any constant, determined by initial conditions.
impulse response
The concept of impulse response comes into play when we consider how a system reacts to an instantaneous impulse, such as the Dirac delta function. An impulse response \( y_p(t) = \delta(t-a) \) in the context of our original differential equation, shows the system's reaction when hit by a sudden spike at \( t = a \). To incorporate this into our solution, a particular solution corresponding to the Dirac impulse must be established. At \( t = a \), the response \( y_p(t) \) is designed to mirror the effect of the impulse:
- Integrate the equation across a small interval around \( t = a \) to find the sudden jump it creates.
- The solution accounts for this jump with \( y(a+\epsilon) = y(a-\epsilon) + 1 \), meaning the system output jumps by 1 unit.
first-order linear differential equation
First-order linear differential equations are the simplest type of differential equations involving a derivative to the first power. They can be written in the standard form: \( \frac{dy}{dt} + p(t) y = g(t) \). These equations are characterized by:
- A linear operation on a function \( y \) and its derivative.
- The coefficient \( p(t) \), which can vary with time, but the function \( y \) and its derivatives only appear to the first power.
- Finding the homogeneous solution, which forms a basis for the general solution, using techniques such as the method of integrating factors.
- Considering particular solutions that capture specific non-homogeneous terms like external impulses or forcing functions.