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HOMEWORK14: Problem 9 Previous Problem Problem List Next Problem (1 point) Use the Laplace transform to solve the following initial value problem: $y'' - 2y = 0$, y(0) = 2, y'(0) = -2 (1) First, using Y for the Laplace transform of y(t), i.e., $Y = \mathcal{L}\{y(t)\}$, find the equation you get by taking the Laplace transform of the differential equation to obtain $(s^2 - 2s)Y(s) + 2s - 6 = 0$ (2) Next solve for Y = (3) Now write the above answer in its partial fraction form, $Y = \frac{A}{(s - a)} + \frac{B}{(s - b)}$ (NOTE: the order that you enter your answers matter so you must order your terms so that the first corresponds to a and the second to b, where a < b. Also note, for example that -2 < 1) Y = (4) Finally apply the inverse Laplace transform to find y(t) y(t) =

          HOMEWORK14: Problem 9
Previous Problem Problem List Next Problem
(1 point)
Use the Laplace transform to solve the following initial value problem: $y'' - 2y = 0$,
y(0) = 2, y'(0) = -2
(1) First, using Y for the Laplace transform of y(t), i.e., $Y = \mathcal{L}\{y(t)\}$,
find the equation you get by taking the Laplace transform of the differential equation to obtain
$(s^2 - 2s)Y(s) + 2s - 6 = 0$
(2) Next solve for Y = 
(3) Now write the above answer in its partial fraction form, $Y = \frac{A}{(s - a)} + \frac{B}{(s - b)}$
(NOTE: the order that you enter your answers matter so you must order your terms so that the first corresponds to a and the second to b, where a < b. Also note, for example that
-2 < 1)
Y = 
(4) Finally apply the inverse Laplace transform to find y(t)
y(t) =
        
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HOMEWORK14: Problem 9
Previous Problem Problem List Next Problem
(1 point)
Use the Laplace transform to solve the following initial value problem: y” - 2y = 0,
y(0) = 2, y'(0) = -2
(1) First, using Y for the Laplace transform of y(t), i.e., Y = â„’{y(t)},
find the equation you get by taking the Laplace transform of the differential equation to obtain
(s^2 - 2s)Y(s) + 2s - 6 = 0
(2) Next solve for Y = 
(3) Now write the above answer in its partial fraction form, Y = (A)/((s - a)) + (B)/((s - b))
(NOTE: the order that you enter your answers matter so you must order your terms so that the first corresponds to a and the second to b, where a < b. Also note, for example that
-2 < 1)
Y = 
(4) Finally apply the inverse Laplace transform to find y(t)
y(t) =

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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HOMEWORK 14: Problem 9 (1 point) Use the Laplace transform to solve the following initial value problem: y'' - 2y' = 0 y(0) = 2, y'(0) = 2 First, using Y for the Laplace transform of y(t), i.e. Y = L{y(t)}, find the equation you get by taking the Laplace transform of the differential equation to obtain: (s^2 - 2s)Y(s) + 2s - 6 Next, solve for Y: Y = A/(s - a) + B/(s - b) Now write the above answer in its partial fraction form, Y: Y = A/(s - a) + B/(s - b) NOTE: The order that you enter your answers matters, so you must order your terms so that the first corresponds to a and the second to b, where a < b. Also note, for example, that 2 < 1. Finally, apply the inverse Laplace transform to find y(t): y(t) =
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-
00:01 Hello student here in this question.
00:02 Consider part 1.
00:03 Here we have given initial value problem that is y double dash minus 4 y is equals to 0.
00:08 Now apply laplace transformation on both side.
00:11 Therefore we'll get laplace of y double dash minus 4 into laplace of y is equals to laplace of 0 that is equals to s square into laplace of y minus s into y of 0 minus s into y dash at 0 minus 4 into laplace of y.
00:30 That is equals to 0 also here we have given the initial condition that is y at 0 is equal to 4 and y dash at 0 is also given is 4 therefore we'll get s squared minus 4 into laplace so 5 of t is equals to y minus 4 is minus 4 is equal to 0 now consider part 2 here from part a we see that s square minus 4 into y is equals to 4 s plus 4 that implies y is equals to 4 s plus 4 divided by s squared minus 4 therefore y can be written as 4 x divided by s minus 2 into s plus 2 plus 4 divided by s minus 2 into s plus 2 now consider part 3 here from part 2 y can be written as 4 s over s -minus 2 into s plus 2 plus 4 over s minus 2 into s plus 2 therefore, using partial fraction y can be written as 3 over s minus 2 plus 2 plus 2 plus 1 over s plus 2 plus 1 over s minus minus 1 over s plus 2 therefore in partial fraction form y can be written as 3 over s minus 2 plus 1 over s plus 2...
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