Using Laplace transform- solved2ydt2-4y - 12tGiven that y - 0 and dy-dt - 9 at t - 0-

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Using Laplace...

Question

Using Laplace transform, solve
$\frac{d^{2} y}{d t^{2}} + 4 y = 12 t$
Given that y = 0 and dy/dt = 9 at t = 0.

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Solution

The correct option is D d
$\frac{d^{2} y}{d t^{2}} + 4 y = 12 t$
By taking Laplace transform of LHS and RHS and using the property of derivatives of transform $L [f^{n} (t)] = s^{n} F (S) - s^{n - 1} f (0) - s^{n - 2} f^{'} (0) . . . . . . f^{n - 1} (0)$ we get subsidiary equation as :
$(s^{2} + 4) Y - s y (0) - y^{'} (0 = \frac{12}{s^{2}}$
$(s^{2} + 4) Y = \frac{12}{s^{2}} + 5 (0) + 9$
$Y = \frac{12}{s^{2} (s^{2} + 4)} + \frac{9}{s^{2} + 4}$
$Y = \frac{3}{s^{2}} + \frac{3}{s^{2} + 4} + \frac{9}{s^{2} + 4}$
$Y = \frac{3}{s^{2}} + \frac{6}{s^{2} + 4}$
$y (t) = L^{- 1} (Y)$
$y (t) = L^{- 1} (\frac{3}{s^{2}}) + L^{- 1} (\frac{6}{s^{2} + 2^{2}})$
$y (t) = 3 t + 3 sin 2 t$

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Using Laplace transform, solve d2ydt2+4y=12t Given that y = 0 and dy/dt = 9 at t = 0.

Using Laplace transform, solve
$\frac{d^{2} y}{d t^{2}} + 4 y = 12 t$
Given that y = 0 and dy/dt = 9 at t = 0.