The Laplace transform of the given initial-value problem
is mathematically given as
![y(t)=\frac{1}{9} e^{4 t}+\frac{17}{9} e^{-5 t}](https://tex.z-dn.net/?f=y%28t%29%3D%5Cfrac%7B1%7D%7B9%7D%20e%5E%7B4%20t%7D%2B%5Cfrac%7B17%7D%7B9%7D%20e%5E%7B-5%20t%7D)
<h3>What is the Laplace transform of the given initial-value problem? y' 5y = e4t, y(0) = 2?</h3>
Generally, the equation for the problem is mathematically given as
![&\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}](https://tex.z-dn.net/?f=%26%5Ctext%20%7B%20Sol%3A-%20%7D%20%5Cquad%20y%5E%7B%5Cprime%7D%2Bs%20y%3De%5E%7B4%20t%7D%2C%20y%280%29%3D2%20%5C%5C%5C%5C%26%5Ctext%20%7B%20Taking%20Laplace%20transform%20of%20%281%29%20%7D%20%5C%5C%5C%5C%26%5Cquad%20L%5Cleft%5By%5E%7B%5Cprime%7D%2B5%20y%5Cright%5D%3D%5Cleft%5B%5Cleft%5Be%5E%7B4%20t%7D%5Cright%5D%5Cright.%20%5C%5C%5C%5C%26%5CRightarrow%20%5Cquad%20L%5Cleft%5By%5E%7B%5Cprime%7D%5Cright%5D%2B5%20L%5By%5D%3D%5Cfrac%7B1%7D%7Bs-4%7D%20%5C%5C%5C%5C%26%5CRightarrow%20%5Cquad%20s%20y%28s%29-y%280%29%2B5%20y%28s%29%3D%5Cfrac%7B1%7D%7Bs-4%7D%20%5C%5C%5C%5C%26%5CRightarrow%20%5Cquad%28s%2B5%29%20y%28s%29%3D%5Cfrac%7B1%7D%7Bs-4%7D%2B2%20%5C%5C%5C%5C%26%5CRightarrow%20%5Cquad%20y%28s%29%3D%5Cfrac%7B1%7D%7Bs%2B5%7D%5Cleft%5B%5Cfrac%7B1%7D%7Bs-4%7D%2B2%5Cright%5D%3D%5Cfrac%7B2%20s-7%7D%7B%28s%2B5%29%28s-4%29%7D%5Cend%7Baligned%7D)
![\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}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%26%5Ctext%20%7B%20Let%20%7D%20%5Cfrac%7B2%20s-7%7D%7B%28s%2B5%29%28s-4%29%7D%3D%5Cfrac%7Ba_%7B0%7D%7D%7Bs-4%7D%2B%5Cfrac%7Ba_%7B1%7D%7D%7Bs%2B5%7D%20%5C%5C%26%5CRightarrow%202%20s-7%3Da_%7B0%7D%28s%2Bs%29%2Ba_%7B1%7D%28s-4%29%5Cend%7Baligned%7D)
![put $s=-s \Rightarrow a_{1}=\frac{17}{9}$](https://tex.z-dn.net/?f=put%20%24s%3D-s%20%5CRightarrow%20a_%7B1%7D%3D%5Cfrac%7B17%7D%7B9%7D%24)
![\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}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%5Ctext%20%7B%20put%20%7D%20s%20%26%3D4%20%5CRightarrow%20a_%7B0%7D%3D%5Cfrac%7B1%7D%7B9%7D%20%5C%5C%5CRightarrow%20%5Cquad%20y%28s%29%20%26%3D%5Cfrac%7B1%7D%7B9%28s-4%29%7D%2B%5Cfrac%7B17%7D%7B9%28s%2Bs%29%7D%5Cend%7Baligned%7D)
In conclusion, Taking inverse Laplace tranoform
![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]$ \\\\](https://tex.z-dn.net/?f=L%5E%7B-1%7D%5By%28s%29%5D%3D%5Cfrac%7B1%7D%7B9%7D%20L%5E%7B-1%7D%5Cleft%5B%5Cfrac%7B1%7D%7Bs-4%7D%5Cright%5D%2B%5Cfrac%7B17%7D%7B9%7D%20L%5E%7B-1%7D%5Cleft%5B%5Cfrac%7B1%7D%7Bs%2B5%7D%5Cright%5D%24%20%5C%5C%5C%5C)
![y(t)=\frac{1}{9} e^{4 t}+\frac{17}{9} e^{-5 t}](https://tex.z-dn.net/?f=y%28t%29%3D%5Cfrac%7B1%7D%7B9%7D%20e%5E%7B4%20t%7D%2B%5Cfrac%7B17%7D%7B9%7D%20e%5E%7B-5%20t%7D)
Read more about Laplace tranoform
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Answer:
9. 81.3953% increase
10. 26.6667% decrease
Step-by-step explanation:
Hope this helps.
the third and forth option:
9.2 × 10-9
3.4 × 10-10
14 is the answer 2cups in 1pint
<span>wavelength = 13 mm
frequency = 14 hertz.
Speed = 13*14 = 182 mm/s
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