Question

Use the Laplace transform to solve the given initial-value problem

$\mathbf{2}\frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{y}\left(0\right)\mathbf{=}\mathbf{-}\mathbf{3}$

Short answer

The Laplace transform of the given derivative equation$2\frac{\mathrm{dy}}{\mathrm{dt}}+\mathrm{y}=0,\mathrm{y}\left(0\right)=-3$is$\mathrm{Y}\left(\mathrm{t}\right)=-3{\mathrm{e}}^{\raisebox{1ex}{$-\mathrm{t}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

Step-by-step solution

Theorem of Transform of a Derivative

If $\mathbf{f}\mathbf{,}{\mathbf{f}}^{\mathbf{\text{'}}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\mathbf{f}}^{\left(n-1\right)}$are continuous on $\mathbf{[}\mathbf{0}\mathbf{,}\mathbf{\infty }\mathbf{)}$and are of exponential order and if ${\mathbf{f}}^{\left(n\right)}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$is a piecewise continuous on $\mathbf{[}\mathbf{0}\mathbf{,}\mathbf{\infty }\mathbf{)}$, then

$\mathbf{L}\left\{f^{\left(n\right)}\left(t\right)\right\}\mathbf{=}{\mathbf{s}}^{\mathbf{n}}\mathbf{F}\left(s\right)\mathbf{-}{\mathbf{s}}^{\mathbf{n}\mathbf{-}\mathbf{1}}\mathbf{f}\left(0\right)\mathbf{-}{\mathbf{s}}^{\mathbf{n}\mathbf{-}\mathbf{2}}{\mathbf{f}}^{\mathbf{\text{'}}}\left(0\right)\mathbf{-}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{-}{\mathbf{f}}^{\left(n-1\right)}\left(0\right)$

Where role="math" localid="1664105203612" $\mathbf{F}\left(s\right)\mathbf{=}\mathbf{L}\left\{f\left(t\right)\right\}$.

The given derivative equation is $2\frac{\mathrm{dy}}{\mathrm{dt}}+\mathrm{y}=0,\mathrm{y}\left(0\right)=-3$

Using the Theorem of Transform of a Derivative, we are transforming the above given derivative,

$$\begin{gathered} L\left\{2y^{\text{'}}+y\right\}=L\left\{0\right\} \\ 2sY-2\left(-3\right)+Y=0\to Y=-\frac{6}{2s+1} \\ =-\frac{1}{2}\times \frac{6}{s-\left(-\frac{1}{2}\right)} \end{gathered}$$

Finding the inverse Transform

Now we are going to find the inverse transform,

$$\begin{gathered} -\frac{6}{2}{\mathrm{L}}^{-1}\times \frac{1}{\mathrm{s}-\left(-\frac{1}{2}\right)} \\ \mathrm{Y}\left(\mathrm{t}\right)=-3{\mathrm{e}}^{\raisebox{1ex}{$-\mathrm{t}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \end{gathered}$$

Therefore, the solved problem is$\mathrm{Y}\left(\mathrm{t}\right)=-3{\mathrm{e}}^{\raisebox{1ex}{$-\mathrm{t}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$.