Solve the initial value problem {y+16y=cos⁡(4t)y(0)=1y′(0)=1 using the Laplace transform.

Step 1Given: y+16y=cos⁡(4t),y(0)=1,y′(0)=1Step 2Explanation:y+16y=cos⁡(4t),y(0)=1,y′(0)=1Taking Laplace transform on both sidesL[y″+16y]=L[cos⁡(4t)]L(y(t))+L(16y(t))=L[cos⁡(4t)]…(i)We know that if L(y(t))=y(s)then L(y″(t))=s2y(s)−sy(0)−y′(0) L[cos⁡(at)]=ss2+a2Using these in equation (i), we gets2y(s)−sy(0)−y′(0)+16y(s)=ss2+42Using the given initial conditions y(0)=1,y′(0)=1 then(s2+16)y(s)−s(1)−1=ss2+42(s2+42)y(s)=1+s+ss2+42y(s)=1s2+42+ss2+42+s(s2+42)(s2+42)y(s)=1s2+42+ss2+42+s(s2+42)2Taking inverse Laplace transform on both sidesL−1(y(s))=L−1(1s2+42+ss2+42+s(s2+42)2)

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Wierzycaz

Answered question

2020-10-23

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 & Explanation

Obiajulu

Skilled2020-10-24Added 98 answers

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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