Question

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

Answer 1

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:

Apply Laplace transform

• $\mathcal{L}\lbrace y(t) \rbrace = Y(s)$.
• $\mathcal{L}\left\lbrace y^\prime \right\rbrace = s\, Y(s) - y(0)$.
• $\begin{aligned}\mathcal{L}\left\lbrace y^{\prime\prime} \right\rbrace &= s\, \mathcal{L}\left\lbrace y^{\prime} \right\rbrace - y^\prime(0) \\ &= s\, \left( s\, Y(s) - y(0) \right) - y^\prime(0) = s^2\, Y(s) - s\cdot y(0) - y^\prime(0) \end{aligned}$.

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:

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

Solve for Y(s)

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:

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

Invert Laplace Transform

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 $\omega = 1$.

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