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Emily-Jane Bray

Answered question

2021-02-09

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

2 t y^{″} - 3 t y^{'} + 3 y = 6, y (0) = 2, y^{'} (0) = - 3

Answer & Explanation

unett

Skilled2021-02-10Added 119 answers

Step 1
Given initial value problem,

2 t y^{″} - 3 t y^{'} + 3 y = 6, y (0) = 2, y^{'} (0) = - 3

Step 2
Take Laplace transform of both sides of the given initial value problem,

L [2 t y^{″} - 3 t y^{'} + 3 y] = L [6]

2 L [t y^{″}] - 3 L [t y^{'}] + 3 L [y] = 6 L [1]

Use the formula such that

L [t y^{″}] = (- 1) \frac{d}{(d s)} L [y^{″}]

L [y^{″}] = s^{2} L [y] - s y (0) - y^{'} (0)

L [y^{'}] = s L [y] - y (0)

L [1] = \frac{1}{s}

Step 3
Then,

2 [s^{2} L [y] - s y (0) - y^{'} (0)] - 3 [s L [y] - y (0)] + 3 L [y] = 6 \cdot \frac{1}{s}

2 [s^{2} L [y] - 2 s + 3] - 3 [s L [y] - 2] + 3 L [y] = \frac{6}{s}

2 s 2 L [y] - 4 s + 6 - 3 s L [y] + 6 + 3 L [y] = \frac{6}{s}

(2 s^{2} - 3 s + 3) L [y] - 4 s + 12 = \frac{6}{s}

(2 s^{2} - 3 s + 3) L [y] = \frac{6}{s} + 4 s - 12

L [y] = \frac{6 + 4 s^{2} - 12 s}{2 s^{2} - 3 s + 3}

L [y] = \frac{6 + 4 s^{2} - 12 s}{2 s^{2} - 3 s + 3}

Step 4
Taking inverse Laplace transform,

y = L^{- 1} [\frac{6 + 4 s^{2} - 12 s}{2 s^{2} - 3 s + 3}] \dots (1)

Now

\frac{6 + 4 s^{2} - 12 s}{2 s^{2} - 3 s + 3} = 2 - \frac{6 s}{2 s^{2} - 3 s + 3}

= 2 - \frac{3 s}{s^{2} - \frac{3}{2 s} + \frac{3}{2}}

= 2 - \frac{3 (s - \frac{3}{4} + \frac{3}{4})}{(s - \frac{3}{4})^{2} + \frac{15}{16}}

= 2 - \frac{3 (s - \frac{3}{4})}{(s - \frac{3}{4})^{2} + \frac{15}{16}} - \frac{9}{4} \frac{1}{(s - \frac{3}{4})^{2} + \frac{15}{16}}

Step 5
Then,

L^{- 1} [\frac{6 + 4 s^{2} - 12 s}{2 s^{2} - 3 s + 3}] = L^{- 1} [2 - \frac{3 (s - \frac{3}{4})}{(s - \frac{3}{4})^{2} + \frac{15}{16}} - \frac{9}{4} \frac{1}{(s - \frac{3}{4})^{2} + \frac{15}{16}}]

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