Chapter 8: Q22P (page 439)

Obtain $L (t e^{- a t} c o s b t)$

Short Answer

Expert verified

Answer

The solution is $L \{t e^{- a t} \cos b t\} - \frac{{(p + a)}^{2} - b^{2}}{{({(φ + a)}^{2} + b^{2})}^{2}}$

Step by step solution

Given information

The given function is $f (t) = t e^{- a t} \cos b t$ and to prove is $L (t e^{a t} \cos b t)$

Definition of Laplace Transformation

A transformation of a function f(x) into the function g(t) that is useful especially in reducing the solution of an ordinary linear differential equation with constant coefficients to the solution of a polynomial equation.

The inverse Laplace transform of a function F(s) is the piecewise-continuous and exponentially-restricted real function f(t)

Properties used for given function

$L (29) : L \{e^{- a t} g (t)\} = G (p + a) L (12) : L \{t \cos a t\} = \frac{p^{2} - a^{2}}{{(p^{2} + a^{2})}^{2}}$

Proof for given function

From L29

$L \{e^{- a i_{g}} g (t)\} = G (p + a)$

Since from L11

$L \{e^{- a i_{g}} g (t)\} = G (p + a)$

Consider $g (t) = t \cos b t$ and by use of (L29), L $\{e^{- a t} g (t)\} = G (p + a); L \{t e^{- w} \cos b t\}$ can be calculated by replace $p b y (p + a)$ in equation as,

$L \{t \cos b t\} = \frac{p^{2} - b^{2}}{{(p^{2} + b^{2})}^{2}} L \{t e^{- a t} \cos b t\} = \frac{{(p + a)}^{2} - b^{2}}{{({(p + a)}^{2} + b^{2})}^{2}}$

Hence, $L \{t_{e}^{- a t} \cos b t\} = \frac{{(p + a)}^{2} - b^{2}}{{({(p + a)}^{2})}^{2}}$

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Obtain $L (t e^{- a t} c o s b t)$

Short Answer

Step by step solution

Given information

Definition of Laplace Transformation

Properties used for given function

Proof for given function

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Obtain L(te-atcosbt)

Short Answer

Step by step solution

Given information

Definition of Laplace Transformation

Properties used for given function

Proof for given function

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Most popular questions from this chapter

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Obtain $L (t e^{- a t} c o s b t)$