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Chapter 8: Q2P (page 448)

Use L34 and L2 to find the inverse transform of $G (p) H (p)$ whenand $G (p) = 1 / (p + a) and H (p) = 1 / (p + b)$ ; your result should be L7 .

Short Answer

Expert verified

The inverse transforms of is $G (p) H (p) i s \frac{1}{b - a} (e^{- a t} - e^{- b 蟿}) .$

Step by step solution

01

Given information.

The given expressions are and $G (p) H (p) i s \frac{1}{(p + a)} a n d H (p) i s \frac{1}{(p + b)}$ .

02

Inverse transform.

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

$L {f} (s) = L {\{f\}} (s) = F (s)$

where L is the Laplace transform.

03

Find the inverse transform of G(p)H(p). .

Calculate Laplace inverse.

$L_{- 1} (G (p)) H (p) = g * h = {鈭�}_{0}^{t} g (t - 蟿) h (蟿) d 蟿$

Calculate Laplace inverse of $G (p) .$

role="math" localid="1659264162995" $L^{- 1} G (p) = L^{- 1} (\frac{1}{p + a}) g (t) = e^{- a t} g (t - 蟿) = e^{- a (t - 蟿)}$

Calculate Laplace inverse of H(p).

$L^{- 1} H (p) = L^{- 1} (\frac{1}{p + b}) h (t) = e^{- b t} h (t - 蟿) = e^{- b 蟿}$

Calculate Laplace inverse of G(p)H(p) .

$L^{- 1} [(\frac{1}{p + a}) (\frac{1}{p + b})] = \frac{1}{b - a} (e^{- a t} - e^{- b 蟿})$

Thus, the Laplace inverse of is $G (p) H (p) i s \frac{1}{b - a} (e^{- a t} - e^{- b 蟿}) .$ .

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Most popular questions from this chapter

(a) Show that $D (e^{ax} y) = e^{ax} (D + a) y, D^{2} (e^{ax} y) = e^{ax} {(D + a)}^{2} y$ , and so on; that is, for any positive integral $n$ , $D^{n} (e^{ax} y) = e^{ax} {(D + a)}^{n} y .$

Thus, show that ifis any polynomial in the operator $D$ , then $L (D) (e^{ax} y) = e^{ax} L (D + a) y$ .

This is called the exponential shift.

(b) Use to show that ${(D - 1)}^{3} (e^{x} y) = e^{x} D^{3} y, (D^{2} + D - 6) (e^{- 3 x} y) = e^{- 3 x} (D^{2} - 5 D) y .$ .

(c) Replace $D$ by $D - a$ , to obtain $e^{ax} P (D) y = P (D - a) e^{ax} y$

This is called the inverse exponential shift.

(d) Using (c), we can change a differential equation whose right-hand side is an exponential times a

polynomial, to one whose right-hand side is just a polynomial. For example, consider

$(D^{2} - D - 6) y = 10 脳 e^{2 x}$ ; multiplying both sides by $e^{- 3 x}$ and using (c), we get

$e^{- 3 x} (D^{2} - D - 6) y = {[{(D + 3)}^{2} - (D + 3) - 6^{"}]}^{"} {ye}^{- 3 x} = (D^{2} + 5 D) {ye}^{- 3 x} = 10 x$

Show that a solution of $(D^{2} + 5 D) u = 10 x$ is $u = x^{2} - \frac{2}{5} x$ ; then or use this method to solve Problems 23 to 26.

Using $(3.9)$ , find the general solution of each of the following differential equations. Compare a computer solution and, if necessary, reconcile it with yours. Hint: See comments just after $(3.9)$ , and Example 1.

$y' + y t a n h x = 2 e^{x}$

The momentum pof an electron at speednear the speedof light increases according to the formula $p = \frac{mv}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}$ , whereis a constant (mass of the electron). If an electron is subject to a constant force F, Newton鈥檚 second law describing its motion is localid="1659249453669" $\frac{d p}{d x} = \frac{d}{d x} \frac{m v}{\sqrt{1 - \frac{v^{2}}{c^{2}}}} = F .$

Find $v (t)$ and show that $v 鈫� c$ as $t 鈫� 鈭�$ . Find the distance travelled by the electron in timeif it starts from rest.

Solve Example 4 using the general solution $y = a \sinh x + b \cosh x$ .

For each of the following differential equations, separate variables and find a solution containing one arbitrary constant. Then find the value of the constant to give a particular solution satisfying the given boundary condition. Computer plot a slope field and some of the solution curves.

$12 . (x + xy) y' + y = 0$ y = 1When x = 1.

See all solutions

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