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x^{1}+y=t16 x+y^{\prime}=0Taking laplace trabform of (1)x(0)=3, \quad y(0)=4\begin{array}{l}S L(x)-y^{\prime} x(0)+L(y)=\frac{1}{s^{2}} \\=S L(x)+L(y)=\frac{1}{s^{2}}+3\end{array}Taking Laplace trampom of (11)\begin{array}{l}16 L(x)+5 \cdot L(y)-y(0)=0 \\160-L(x)+5 \cdot L(y)=4\end{array}frow (III) \infty (w)\begin{array}{l}S^{2} L(x)+S L(y)=\frac{1}{s}+3 s \\16 L(x)+S L(y)=-1 \\\left(s^{2}-16\right) L(x)=\frac{1}{s}+3 s-4 \\\Rightarrow L(x)=\frac{1}{s\left(s^{2}-16\right)}+\frac{3 s}{s^{2}-16}-\frac{4}{s^{2}-16} \\\end{array}Takive inverse Laplace tramform we haex(t)=L^{-1}\left\{\frac{1}{s\left(s^{2}-16\right.}\right\}+3 \cdot L^{-1}\left\{\frac{s}{s^{2}-16}\right\}-L^{-1}\left(\frac{4}{s^{2}-16}\right)row\begin{aligned}\frac{1}{s \cdot\left(s^{2}-16\right)} & =\frac{1}{32(s+4)}+\frac{1}{32(s-4)}-\frac{1}{16 s} \\\frac{3 s}{s}-16 & =\frac{3}{2(s+4)}+\frac{3}{2(s-4)} \\\frac{4}{s^{2}-16} & =-\frac{1}{2(s+4)}+\frac{1}{2(s-4)}\end{aligned}\begin{array}{l}\begin{array}{l}\therefore \frac{1}{s\left(s^{2}-16\right)}+\frac{3 s}{s^{2}-16}-\frac{4}{s^{2}-16}=\frac{65}{32}+\frac{1}{s+4}+\frac{33}{32} \frac{1}{s+4} \\L^{-1}\left\{\frac{1}{s\left(s^{2}-16\right)}+\frac{3 s}{s^{2}-16}-\frac{4}{s^{2}-16}\right\}=\frac{65}{32} e^{-4 t}+\frac{33}{32} e^{4 t}-\frac{1}{16} \\x(1)-4 t\end{array} \\-x(t)=\frac{65}{32} e^{-4 t}+\frac{33}{32} e^{4 t}+\frac{1}{16} \text {. } \\\end{array}\begin{array}{l}x^{\prime}(t)=-\frac{65}{8} e^{-4 t}+\frac{33}{8} e^{4 t} \\y(t)=t-x^{\prime}(t)=t+\frac{65}{8} e^{-4 t}-\frac{33}{8} e^{4 t}\end{array}\angle y(t)=t+\frac{65}{8} e^{-4 t}-\frac{33}{8} e^{4 t}. ...