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Chapter 8: Q11-17P (page 460)

Verify (11.18a) and (11.18c) by multiplying by a test function and integrating.

Short Answer

Expert verified

The test function is $x^{2} δ^{'} (x) = 2 δ (x)$ .

Step by step solution

01

Given information

The given expression is

$\int ϕ (x) δ (x) d x$

02

Definition of Integration by Parts

Integration by partsor partial integration is a process that finds theintegralof aproductoffunctionsin terms of the integral of the product of theirderivativeandanti-derivative.

03

Verify the Test Function

Consider

$\int ϕ (x) δ (x) d x = \int (ϕ (x)) δ (x) d x = (x ϕ (x))_{ats = 0}$

Hence $x ϕ (x) = 0$

Consider

$\int_{- \infty}^{\infty} ϕ (x) x^{2} δ (x)^{''} d x = \int_{- \infty}^{\infty} 2 ϕ (x) d x + \int_{- \infty}^{\infty} 4 x ϕ (x)^{'} d x + \int_{- \infty}^{\infty} x^{2} ϕ (x)^{''} d x = {2 ϕ (x)|}_{x = 0} + {4 x ϕ {(x)}^{'}|}_{x = 0} + {x^{2} ϕ {(x)}^{''}|}_{x = 0} = 2 ϕ (0) = \int_{- \infty}^{\infty} ϕ (x) 2 δ (x) d x$

Hence, $x^{2} δ^{'} (x) = 2 δ (x)$

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Most popular questions from this chapter

By using Laplace transforms, solve the following differential equations subject to the given initial conditions. $y^{'}' + y^{'} - 5 y = e^{2} t, y_{0} = 1, y_{0}^{'} = 2$

Sketch on the same axes graphs of $s i n t, s i n (t - π / 2)$ , and $s i n (t + π / 2)$ , and observe which way the graph shifts. Hint: You can, of course, have your calculator or computer plot these for you, but it's simpler and much more useful to do it in your head. Hint: What values of $t$ make the sines equal to zero? For an even simpler example, sketch on the same axes $y = t, y = t - π / 2, y = t + π / 2$ .

By using Laplace transforms, solve the following differential equations subject to the given initial conditions.

$y " - 4 y' + 4 y = 6 e^{2 t}, y_{0} = y^{'} = 0$

Use L29 and L11 to obtain $L (t e^{- a t} s i n b t)$ which is not in the table.

Use the methods of this section to solve the following differential equations. Compare computer solutions and reconcile differences.

$y' = c o s (x + y)$

Hint: Let $u = x + y$ ; then $u' = 1 + y'$ .

See all solutions

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