Chapter 7: Q17E (page 404)

In problem 15-22, solve the given integral equation or integro differential equation for $y (t)$
$L \int_{0}^{t} (t - v) y (v) d v = 1$

Short Answer

Expert verified

The integral equation or integro-differential equation for y(t)is. $y (t) = \cos t$

Step by step solution

Definition of Laplace transform

A transformation of a function f(x) into the function is particularly useful for reducing the solution of an ordinary linear differential equation with constant coefficients to the solution of a polynomial equation.

Applying the Laplace transform we get

$\begin{matrix} L {y (t) + \int_{0}^{t} (t - v) y (v) d v} (s) = {1} \\ L {y (t)} (s) + L {\int_{0}^{t} (t - v) y (v) d v} (s) = \frac{1}{s} \end{matrix}$

$\begin{matrix} Y (s) = L {t * y (t)} (s) \\ = \frac{1}{s} \end{matrix}$

$\begin{matrix} Y (s) = L {t} (s) . L {y (t)} \\ = \frac{1}{s} \end{matrix}$

$\begin{matrix} Y (s) = 1 + \frac{1}{s^{2}} Y (s) \\ = \frac{1}{s} \end{matrix}$

$\begin{matrix} (1 + \frac{1}{s^{2}}) Y (s) = \frac{1}{s} \\ Y (s) = \frac{s}{S (s^{2} + 1)} \end{matrix}$

Definition of inverse

Now,

using inverse Laplace we get

$\begin{matrix} y (t) = L^{- 1} {\frac{s}{S (s^{2} + 1)}} \\ = \cos t \end{matrix}$

Hence,

$y (t) = \cos t$

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In problem 15-22, solve the given integral equation or integro differential equation for $y (t)$
$L \int_{0}^{t} (t - v) y (v) d v = 1$

Short Answer

Step by step solution

Definition of Laplace transform

Definition of inverse

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In problem 15-22, solve the given integral equation or integro differential equation fory(t)L∫0t(t−v)y(v)dv=1

Short Answer

Step by step solution

Definition of Laplace transform

Definition of inverse

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Decision Maths

Calculus

Logic and Functions

Probability and Statistics

Geometry

Mechanics Maths

Study anywhere. Anytime. Across all devices.

In problem 15-22, solve the given integral equation or integro differential equation for $y (t)$
$L \int_{0}^{t} (t - v) y (v) d v = 1$