Question

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

${\mathbf{y}}^{\mathbf{\text{'}}}\mathbf{-}\mathbf{y}\mathbf{=}\mathbf{2}\mathbf{cos}\mathbf{5}\mathbf{t}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$

Short answer

The Laplace transform of the given derivative equation${\mathrm{y}}^{\text{'}}-\mathrm{y}=2\cos 5\mathrm{t},\mathrm{y}\left(0\right)=0$is$Y\left(t\right)=\frac{5}{13}\sin \left(5t\right)-\frac{1}{13}\cos \left(5t\right)+\frac{1}{13}{e}^t$.

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

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

Where$\mathbf{F}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{=}\mathbf{L}\mathbf{\left\{}\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{\right\}}$

The given derivative equation is $y^{\text{'}}-y=2\cos 5t,y\left(0\right)=0$

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

$$\begin{gathered} L\left\{y^{\text{'}}-y\right\}=2L\left\{\cos 5t\right\} \\ sY-0-Y=\frac{2s}{s^2+25}\to Y=\frac{2s}{\left(s^2+25\right)\left(s-1\right)} \end{gathered}$$

Finding the inverse Transform

Now we need to decompose,

$$\begin{gathered} \frac{2s}{\left(s^2+25\right)\left(s-1\right)}=\frac{as+b}{s^2+25}+\frac{c}{s-1} \\ \frac{2s}{\left(s^2+25\right)\left(s-1\right)}=\frac{25-s}{13\left(s^2+25\right)}+\frac{1}{13\left(s-1\right)} \end{gathered}$$

Now we are going to find the inverse transform,

$$\begin{gathered} \frac{25}{13}{\mathrm{L}}^{-1}\frac{1}{5}\times \frac{5}{{\mathrm{s}}^2+25}-\frac{1}{13}{\mathrm{L}}^{-1}\frac{\mathrm{s}}{{\mathrm{s}}^2+25}+\frac{1}{13}{\mathrm{L}}^{-1}\frac{1}{\mathrm{s}-1} \\ \mathrm{y}\left(\mathrm{t}\right)=\frac{5}{13}\sin \left(5\mathrm{t}\right)-\frac{1}{13}\cos \left(5\mathrm{t}\right)+\frac{1}{13}{\mathrm{e}}^{\mathrm{t}} \end{gathered}$$

Therefore, The Laplace transform of the given derivative equation $y^{\text{'}}-y=2\cos 5t,y\left(0\right)=0$is $Y\left(t\right)=\frac{5}{13}\sin \left(5t\right)-\frac{1}{13}\cos \left(5t\right)+\frac{1}{13}{e}^t$.