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miss Akunina [59]
3 years ago
13

Use the Laplace transform to solve the following initial value problem:

Mathematics
1 answer:
Sindrei [870]3 years ago
4 0

Answer:

Laplace transform of the original differential equation:

\left( s^2\, Y(s) - s\cdot y(0) - y^\prime(0)\right) - 4\, (s\, Y(s) - y(0)) + 5\, Y(s) = 0.

Given that y(0) = 1 and y^\prime(0) = 2, solve for Y(s) to obtain:

\displaystyle Y(s) = \frac{s - 2}{s^2 - 4\, s + 5} = \frac{s - 2}{(s - 2)^2 + 1}.

Apply inverse Laplace transform to obtain:

y(t) = e^{2 t} \, \sin(t).

Step-by-step explanation:

<h3>Apply Laplace transform</h3>
  • \mathcal{L}\lbrace y(t) \rbrace = Y(s).
  • \mathcal{L}\left\lbrace y^\prime \right\rbrace = s\, Y(s) - y(0).
  • \begin{aligned}\mathcal{L}\left\lbrace y^{\prime\prime} \right\rbrace &= s\, \mathcal{L}\left\lbrace y^{\prime} \right\rbrace - y^\prime(0) \\ &= s\, \left( s\, Y(s) - y(0) \right) - y^\prime(0) = s^2\, Y(s) - s\cdot y(0) - y^\prime(0) \end{aligned}.

Apply these two rules to replace all y(t), y^\prime(t), and y^{\prime\prime}(t) in the original equation with their Laplace transforms:

\underbrace{\left( s^2\, Y(s) - s\cdot y(0) - y^\prime(0)\right)}_{\mathcal{L}\left\lbrace y^{\prime\prime}(t) \right\rbrace} - 4\, \underbrace{(s\, Y(s) - y(0))}_{\mathcal{L}\left\lbrace y^{\prime}(t) \right\rbrace} + 5\, \underbrace{Y(s)}_{\mathcal{L}\left\lbrace y(t) \right\rbrace} = 0.

<h3>Solve for Y(s)</h3>

Substitute in the values y(0) = 1 and y^\prime(0) = 2.

{\left( s^2\, Y(s) - s - 2 \right)}  - 4\, (s\, Y(s) - 2) + 5\, Y(s) = 0.

Solve for Y(s) after rearranging this equation:

\displaystyle Y(s) = \frac{s - 2}{s^2 - 4\, s + 5}.

Note that if denominator is the left-hand side of a quadratic equation, this equation would have no real root. Hence, complete the square in the denominator:

\displaystyle Y(s) = \frac{s - 2}{s^2 - 4\, s + 5} = \frac{s - 2}{(s - 2)^2 + 1^2}.

<h3>Invert Laplace Transform</h3>

Look up a table of Laplace transforms. Apply the rule \displaystyle \mathcal{L}\left\lbrace e^{\lambda t}\sin(\omega\, t) \right\rbrace = \frac{s - \lambda}{(s - \lambda)^2 + \omega^2}, where \lambda = 2 and \omega = 1.

y(t) = \mathcal{L} \left\lbrace Y(s) \right\rbrace = e^{2 t}\, \sin(t).

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