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Given the second order initial value problem $y'' = 05(t - 2)$, $y(0) = 4$, $y'(0) = 0$\ Let $Y(s)$ denote the Laplace transform of $y$. Then\ $Y(s) = \frac{4}{s^2}$\ Taking the inverse Laplace transform we obtain\ y(t) =

          Given the second order initial value problem $y'' = 05(t - 2)$, $y(0) = 4$, $y'(0) = 0$\
Let $Y(s)$ denote the Laplace transform of $y$. Then\
$Y(s) = \frac{4}{s^2}$\
Taking the inverse Laplace transform we obtain\
y(t) =
        
Given the second order initial value problem y” = 05(t - 2), y(0) = 4, y'(0) = 0Let Y(s) denote the Laplace transform of y. ThenY(s) = (4)/(s^2)Taking the inverse Laplace transform we obtainy(t) =

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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Given the second order initial value problem y'' + 3ty' - 2y = 4, y(0) = 0. Let Y(s) denote the Laplace transform of y. Then Y(s) = 4/s^2 Taking the inverse Laplace transform, we obtain y(t) =
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Transcript

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00:01 Hello, so we have to solve this question by the laplace formula.
00:07 Writing the differential equation y double dash minus 4y is equal to 4 into delta of t minus 3 given y of 0 as minus 1 and y dash of 0 as 2.
00:24 Now taking laplace of y double dash minus 4y is equal to laplace of 4 into delta of t minus 3.
00:38 Laplace of y double dash minus 4 into laplace of y will be 4 into laplace of delta of t minus 3.
00:49 Y of 0 is minus 1, y dash of 0 is 2.
00:58 So we can here say that now laplace of delta of t minus 3 is e raised to minus 3 is laplace of y of 0 is laplace of minus 1.
01:20 Laplace of y dash of 0 is laplace of 2.
01:31 So y of s will be minus 1 by s.
01:35 S y of s minus y of 0 is it is equal to 2.
01:45 Further we can say that s y of s plus 1 is 2.
01:52 S y of s will be 1 by s.
01:57 So now we will say that s square y of s minus 4 y of s is equal to 4...
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