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Zoe Oneal

Answered question

2020-11-17

Solve the initial value problem below using the method of Laplace transforms

y " - 35 y = 144 t - 36^{- 6 t}

y (0) = 0

y^{'} (0) = 47

Answer & Explanation

Raheem Donnelly

Skilled2020-11-18Added 75 answers

Solution:
The differential equation is

y " - 35 y = 144 t - 36^{- 6 t}

y (0) = 0

y^{'} (0) = 47

Apply Laplace transforms:

L {y " - 35 y} = L {144 t - 36^{- 6 t}}

s^{2} Y (s) - s y (0) - y^{'} (0) - 36 Y (s) = \frac{144}{s^{2}} - \frac{36}{s + 6}

(s^{2} - 36) Y (s) - 47 = \frac{144}{s^{2}} - \frac{36}{s + 6}

Y (s) = \frac{144}{s^{2} (s^{2} - 36)} - \frac{36}{(s + 6) (s^{2} - 36)} + \frac{47}{(s^{2} - 36)}

Decompose each term:

\frac{144}{s^{2} (s^{2} - 36)} = \frac{A}{s} + \frac{B}{s^{2}} + \frac{C}{s + 6} + \frac{D}{s - 6}

= \frac{A s (s^{2} - 36)}{s^{2} (s^{2} - 36)} + \frac{B (s^{2} - 36)}{s^{2} (s^{2} - 36)} + \frac{C s^{2} (s - 6)}{s^{2} (s^{2} - 36)} + \frac{D s^{2} (s + 6)}{s^{2} (s^{2} - 36)}

= \frac{(A + C + D) s^{3} + (B - 6 C + 6 D) s^{2} - 36 A s - 36 B}{s^{2} (s^{2} - 36)}

A = 0, B = - 4, C = - \frac{1}{3}, and D = \frac{1}{3}

Step 2

\frac{144}{s^{2} (s^{2} - 36)} = \frac{- 4}{s^{2}} - \frac{1}{3 (s + 6)} + \frac{1}{3 (s - 6)}

\frac{36}{(s + 6) (s^{2} - 36)} = \frac{A}{s + 6} + \frac{B}{(s + 6)^{2}} + \frac{C}{(s - 6)}

= \frac{A (s^{2} - 36) + B (s - 6) + C (s + 6)^{2}}{(s + 6) (s^{2} - 36)}

A = \frac{1}{4}, B = 3, C = - \frac{1}{4}

\frac{36}{(s + 6) (s^{2} - 36)} = \frac{1}{4 (s + 6)} + \frac{3}{(s + 6)^{2}} - \frac{1}{4 (s - 6)}

\frac{47}{(s^{2} - 36)} = \frac{A}{s + 6} + \frac{B}{s - 6}

A = - \frac{47}{12}, B = \frac{47}{12}

Hence, \frac{47}{(s^{2} - 36)} = - \frac{47}{12 (s + 6)} + \frac{47}{12 (s - 6)}

Conclusion:
Hence, we have

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