The Laplace transform of the given initial-value problem
[tex]y' 5y = e^{4t}, y(0) = 2[/tex] is mathematically given as
[tex]y(t)=\frac{1}{9} e^{4 t}+\frac{17}{9} e^{-5 t}[/tex]
Generally, the equation for the problem is mathematically given as
[tex]&\text { Sol:- } \quad y^{\prime}+s y=e^{4 t}, y(0)=2 \\\\&\text { Taking Laplace transform of (1) } \\\\&\quad L\left[y^{\prime}+5 y\right]=\left[\left[e^{4 t}\right]\right. \\\\&\Rightarrow \quad L\left[y^{\prime}\right]+5 L[y]=\frac{1}{s-4} \\\\&\Rightarrow \quad s y(s)-y(0)+5 y(s)=\frac{1}{s-4} \\\\&\Rightarrow \quad(s+5) y(s)=\frac{1}{s-4}+2 \\\\&\Rightarrow \quad y(s)=\frac{1}{s+5}\left[\frac{1}{s-4}+2\right]=\frac{2 s-7}{(s+5)(s-4)}\end{aligned}[/tex]
[tex]\begin{aligned}&\text { Let } \frac{2 s-7}{(s+5)(s-4)}=\frac{a_{0}}{s-4}+\frac{a_{1}}{s+5} \\&\Rightarrow 2 s-7=a_{0}(s+s)+a_{1}(s-4)\end{aligned}[/tex]
[tex]put $s=-s \Rightarrow a_{1}=\frac{17}{9}$[/tex]
[tex]\begin{aligned}\text { put } s &=4 \Rightarrow a_{0}=\frac{1}{9} \\\Rightarrow \quad y(s) &=\frac{1}{9(s-4)}+\frac{17}{9(s+s)}\end{aligned}[/tex]
In conclusion, Taking inverse Laplace tranoform
[tex]L^{-1}[y(s)]=\frac{1}{9} L^{-1}\left[\frac{1}{s-4}\right]+\frac{17}{9} L^{-1}\left[\frac{1}{s+5}\right]$ \\\\[/tex]
[tex]y(t)=\frac{1}{9} e^{4 t}+\frac{17}{9} e^{-5 t}[/tex]
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