Solve by using Laplace transformy"+y=cos⁡t ,y(0)=0 , y′(0)=−1

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Brittney Lord

Answered question

2020-10-28

Solve by using Laplace transform

y " + y = \cos t, y (0) = 0, y^{'} (0) = - 1

Answer & Explanation

Benedict

Skilled2020-10-29Added 108 answers

Step 1
The given initial value problem is

y " + y = \cos t, y (0) = 0, y^{'} (0) = - 1

Apply Laplace transform on both sides of the given equation as follows.

L {y "} + L {y} = L {\cos t}

s^{2} Y (s) - s y (0) - y^{'} (0) + Y (s) = \frac{s}{s^{2} + 1}

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

Y (s) = \frac{s - s^{2} - 1}{(s^{2} + 1)^{2}}

Step 2

By partial fraction decomposition,

\frac{s - s^{2} - 1}{(s^{2} + 1)^{2}} = \frac{A s + B}{(s^{2} + 1)} + \frac{C s + D}{(s^{2} + 1)^{2}}

= \frac{(A s + B) (s^{2} + 1) + C s + D}{(s^{2} + 1)^{2}}

Then,

s - s^{2} - 1 = (A s + B) (s^{2} + 1) + C s + D

= A s^{3} + A s + B s^{2} + B + C s + D

= A s^{3} + B s^{2} + (A + C) s + (B + D)

Step 3

Equating the coefficients of like terms on both sides,

A = 0

B = - 1

A + C = 1 \Rightarrow 0 + C = 1 \Rightarrow C = 1

B + D = - 1 \Rightarrow D = - 1 + 1 \Rightarrow D = 0

Therefore, we have,

\frac{s - s^{2} - 1}{(s^{2} + 1)^{2}} = \frac{A s + B}{(s^{2} + 1)} + \frac{C s + D}{(s^{2} + 1)^{2}}

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

Step 4

Then,

Y (s) = \frac{s - s^{2} - 1}{(s^{2} + 1)^{2}}

= \frac{A s + B}{(s^{2} + 1)} + \frac{C s + D}{(s^{2} + 1)^{2}}

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

Now take inverse Laplace transform on both sides,

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

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