Answer:
Laplace transform of the original differential equation:
[tex]\left( s^2\, Y(s) - s\cdot y(0) - y^\prime(0)\right) - 4\, (s\, Y(s) - y(0)) + 5\, Y(s) = 0[/tex].
Given that [tex]y(0) = 1[/tex] and [tex]y^\prime(0) = 2[/tex], solve for [tex]Y(s)[/tex] to obtain:
[tex]\displaystyle Y(s) = \frac{s - 2}{s^2 - 4\, s + 5} = \frac{s - 2}{(s - 2)^2 + 1}[/tex].
Apply inverse Laplace transform to obtain:
[tex]y(t) = e^{2 t} \, \sin(t)[/tex].
Step-by-step explanation:
Apply these two rules to replace all [tex]y(t)[/tex], [tex]y^\prime(t)[/tex], and [tex]y^{\prime\prime}(t)[/tex] in the original equation with their Laplace transforms:
[tex]\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[/tex].
Substitute in the values [tex]y(0) = 1[/tex] and [tex]y^\prime(0) = 2[/tex].
[tex]{\left( s^2\, Y(s) - s - 2 \right)} - 4\, (s\, Y(s) - 2) + 5\, Y(s) = 0[/tex].
Solve for [tex]Y(s)[/tex] after rearranging this equation:
[tex]\displaystyle Y(s) = \frac{s - 2}{s^2 - 4\, s + 5}[/tex].
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:
[tex]\displaystyle Y(s) = \frac{s - 2}{s^2 - 4\, s + 5} = \frac{s - 2}{(s - 2)^2 + 1^2}[/tex].
Look up a table of Laplace transforms. Apply the rule [tex]\displaystyle \mathcal{L}\left\lbrace e^{\lambda t}\sin(\omega\, t) \right\rbrace = \frac{s - \lambda}{(s - \lambda)^2 + \omega^2}[/tex], where [tex]\lambda = 2[/tex] and [tex]\omega = 1[/tex].
[tex]y(t) = \mathcal{L} \left\lbrace Y(s) \right\rbrace = e^{2 t}\, \sin(t)[/tex].
