In Problems 1–19, use the method of Laplace transforms to solve the given initial value problem. Here x′, y′, etc., denotes differentiation with respect to t; so does the symbol D.

$\begin{array}{l}\mathbf{D}\mathbf{\left[}\mathbf{x}\mathbf{\right]}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{7}\mathbf{/}\mathbf{4}\mathbf{,}\\ \mathbf{4}\mathbf{x}\mathbf{+}\mathbf{D}\mathbf{\left[}\mathbf{y}\mathbf{\right]}\mathbf{=}\mathbf{3}\mathbf{;}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{4}\end{array}$

The differential equations are given as:

$\begin{array}{l}D\left[x\right]+y=0;x\left(0\right)=7/4,\\ 4x+D\left[y\right]=3;y\left(0\right)=4\end{array}$

Given initial value equations are,

$\begin{array}{l}D\left[x\right]+y=0;x\left(0\right)=7/4,\\ \therefore x^{\text{'}}+y=\mathrm{0....}\left(1\right)\\ 4x+D\left[y\right]=3;y\left(0\right)=4\\ \therefore 4x+y^{\text{'}}=\mathrm{3.....}\left(2\right)\end{array}$

Taking Laplace transform of equation first we get

$\begin{array}{l}sx\left(s\right)-x\left(0\right)+y\left(s\right)=0\\ sx\left(s\right)-\frac{7}{4}+y\left(s\right)=0\\ y\left(s\right)=\frac{7}{4}-sx\left(s\right)\mathrm{.....}\left(3\right)\end{array}$

Taking Laplace transform of equation second we get

$\begin{array}{r}4x\left(s\right)+sy\left(s\right)-y\left(0\right)=\frac{3}{s}\\ 4x\left(s\right)+sy\left(s\right)-4=\frac{3}{s}\mathrm{...}\left(4\right)\end{array}$

Putting equation third into fourth we get

$\begin{array}{l}4x\left(s\right)+s\left[\frac{7}{4}-sx\left(s\right)\right]-4=\frac{3}{s}\\ [4-s^2]x\left(s\right)=\frac{12+16s-7s^2}{4s}\\ x\left(s\right)=\frac{7s^2-16s-12}{4s(s^2-4)}\end{array}$

Using partial fraction we can write as

$=\frac{3}{4s}+\frac{3}{2(s+2)}-\frac{1}{2(s-2)}$

Taking inverse Laplace transform we get

$x\left(t\right)=\frac{3}{4}+\frac{3e^{-2t}}{2}-\frac{1e^{2t}}{2}$

Since equation first is,

$\begin{array}{l}{x}^{\text{'}}+y=0\\ -3e^{-2t}-e^{2t}+y\left(t\right)=0\\ y\left(t\right)=3e^{-2t}+e^{2t}\end{array}$

Hence

$\begin{array}{l}x\left(t\right)=\frac{3}{4}+\frac{3e^{-2t}}{2}-\frac{1e^{2t}}{2},y\left(t\right)=3e^{-2t}+e^{2t}\\ \end{array}$

The final solution is

$\begin{array}{l}x\left(t\right)=\frac{3}{4}+\frac{3e^{-2t}}{2}-\frac{1e^{2t}}{2},y\left(t\right)=3e^{-2t}+e^{2t}\\ \end{array}$