Solve the initial value problem {(y+16y=cos(4t)),(y(0)=1),(y'(0)=1):} using the Laplace transform.

Question

Solve the initial value problem ${\begin{matrix} y + 16 y = \cos (4 t) \\ y (0) = 1 \\ y^{'} (0) = 1 \end{matrix}$ using the Laplace transform.

Answer 1

Step 1
Given:

y + 16 y = \cos (4 t), y (0) = 1, y^{'} (0) = 1

Step 2
Explanation:

y + 16 y = \cos (4 t), y (0) = 1, y^{'} (0) = 1

Taking Laplace transform on both sides

L [y^{″} + 16 y] = L [\cos (4 t)]

L (y (t)) + L (16 y (t)) = L [\cos (4 t)] \dots (i)

We know that if

L (y (t)) = y (s)

then

L (y^{″} (t)) = s^{2} y (s) - s y (0) - y^{'} (0)

L [\cos (a t)] = \frac{s}{s^{2} + a^{2}}

Using these in equation (i), we get

s^{2} y (s) - s y (0) - y^{'} (0) + 16 y (s) = \frac{s}{s^{2} + 4^{2}}

Using the given initial conditions

y (0) = 1, y^{'} (0) = 1

then

(s^{2} + 16) y (s) - s (1) - 1 = \frac{s}{s^{2} + 4^{2}}

(s^{2} + 4^{2}) y (s) = 1 + s + \frac{s}{s^{2} + 4^{2}}

y (s) = \frac{1}{s^{2} + 4^{2}} + \frac{s}{s^{2} + 4^{2}} + \frac{s}{(s^{2} + 4^{2}) (s^{2} + 4^{2})}

y (s) = \frac{1}{s^{2} + 4^{2}} + \frac{s}{s^{2} + 4^{2}} + \frac{s}{{(s^{2} + 4^{2})}^{2}}

Taking inverse Laplace transform on both sides

L^{- 1} (y (s)) = L^{- 1} (\frac{1}{s^{2} + 4^{2}} + \frac{s}{s^{2} + 4^{2}} + \frac{s}{{(s^{2} + 4^{2})}^{2}})

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