Question

Take the Laplace transform of the following initial value and solve for $Y(s) = mathcal{L}{y(t)}$: $y'' + y = egin{cases} sin(pi t), & 0 < t < 1 \ 0, & 1 le t end{cases}$ $y(0) = 0$, $y'(0) = 0$ $Y(s) = Box$. Hint: write the right hand side in terms of the Heaviside function. Now find the inverse transform: $y(t) = Box$ Note: $frac{pi}{(s^2 + pi^2)(s^2 + 1)} = frac{pi}{pi^2 - 1} left(frac{1}{s^2 + 1} - frac{1}{s^2 + pi^2} ight)$ (Notation: write $u(t - c)$ for the Heaviside step function $u_c(t)$ with step at $t = c$.)

          Take the Laplace transform of the following initial value and solve for $Y(s) = mathcal{L}{y(t)}$:
$y'' + y = egin{cases} sin(pi t), & 0 < t < 1 \ 0, & 1 le t end{cases}$ 
$y(0) = 0$, $y'(0) = 0$
$Y(s) = Box$. Hint: write the right hand side in terms of the Heaviside function.
Now find the inverse transform:
$y(t) = Box$
Note:
$frac{pi}{(s^2 + pi^2)(s^2 + 1)} = frac{pi}{pi^2 - 1} left(frac{1}{s^2 + 1} - frac{1}{s^2 + pi^2}
ight)$
(Notation: write $u(t - c)$ for the Heaviside step function $u_c(t)$ with step at $t = c$.)

$Take the Laplace transform of the following initial value and solve for Y(s) = mathcalLy(t): y” + y = egincases sin(pi t), 0 < t < 1 0, 1 le t endcases y(0) = 0, y'(0) = 0 Y(s) = Box. Hint: write the right hand side in terms of the Heaviside function. Now find the inverse transform: y(t) = Box Note: fracpi(s^2 + pi^2)(s^2 + 1) = fracpipi^2 - 1 left(frac1s^2 + 1 - frac1s^2 + pi^2 ight) (Notation: write u(t - c) for the Heaviside step function uc(t) with step at t = c.)$

Added by Andrew P.

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