Using Laplace transform, solve for x(t) for the following system of linear ordinary differential equations dx 4- + 6x + y = cosht dt d²x dt² dy dt with initial conditions x(0) = 0, x'(0) = 0 and y(0) = 0.

Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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hello! please help me solve this! calculus questions. Also, i've provide the appendix for sample equations you to refer!

Using Laplace transform, solve for x(t) for the following system of linear ordinary
differential equations
4
dx
dt
d²x
dt²
+ 6x + y = cosht
+x-
dy
dt
with initial conditions x(0) = 0, x' (0) = 0 and y(0) = 0.
Transcribed Image Text:Using Laplace transform, solve for x(t) for the following system of linear ordinary differential equations 4 dx dt d²x dt² + 6x + y = cosht +x- dy dt with initial conditions x(0) = 0, x' (0) = 0 and y(0) = 0.
Function
a (constant)
f
t"
sin at
cos at
sinh at
cosh at
eat f(t)
f(t-a)u(t-a)
t" f(t)
f(n) (t)
f(u) du
f(u)g(t-u) du
Trigonometric Identities:
APPENDIX
TABLE OF LAPLACE TRANSFORMS
Laplace Transforms
a
S
1
s-a
n!
50+1
a
s² + a²
S
+ a
2
a
s²-a²
S
2
s² - a²
2
F(s-a)
e-as F(s)
(-1)"F(")(s)
s"F(s)-sn-1f(0)-sn-2f (¹) (0) --
F(s)
S
F(s) G(s)
2sinAcosB = sin (A + B) + sin (A - B)
2cosAcosB = cos (A - B) + cos (A + B)
2sinAsinB = cos (A-B)- cos (A + B)
-f(n-1) (0)
sin 2A = 2 sin A cos A
cos 2A = cos²A - sin³A = 2 cos²A-1 = 1-2 sin²A
sin (A + B) = sin A cos B ± cos A sin B
sin A sin B
cos (A + B) = cos A cos B
cosh?A – sinh?A = 1
sinh (A + B) = sinh A cosh B + cosh A sinh B
cosh (A + B) = cosh A cosh B+ sinh A sinh B
Transcribed Image Text:Function a (constant) f t" sin at cos at sinh at cosh at eat f(t) f(t-a)u(t-a) t" f(t) f(n) (t) f(u) du f(u)g(t-u) du Trigonometric Identities: APPENDIX TABLE OF LAPLACE TRANSFORMS Laplace Transforms a S 1 s-a n! 50+1 a s² + a² S + a 2 a s²-a² S 2 s² - a² 2 F(s-a) e-as F(s) (-1)"F(")(s) s"F(s)-sn-1f(0)-sn-2f (¹) (0) -- F(s) S F(s) G(s) 2sinAcosB = sin (A + B) + sin (A - B) 2cosAcosB = cos (A - B) + cos (A + B) 2sinAsinB = cos (A-B)- cos (A + B) -f(n-1) (0) sin 2A = 2 sin A cos A cos 2A = cos²A - sin³A = 2 cos²A-1 = 1-2 sin²A sin (A + B) = sin A cos B ± cos A sin B sin A sin B cos (A + B) = cos A cos B cosh?A – sinh?A = 1 sinh (A + B) = sinh A cosh B + cosh A sinh B cosh (A + B) = cosh A cosh B+ sinh A sinh B
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