The Laplace transform of the given initial-value problem
is mathematically given as
<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)
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)
Read more about Laplace tranoform
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1.63 is the answer. :))))))
Answer:
1) B. similar
2) D. (4, 2)
Step-by-step explanation:
1)
When a figure is dilated, the shapes will be similar. Reflected is another transformation so that's wrong. Same size and congruent is wrong since that would have to make the dilated size equal which can't since it's contradicting.
2)
B: (4, -2)
Reflection across the x-axis:
(x, y) → (x, -y)
(4, -2) → (x, -y)
(4, 2)
Answer:
k=|14+x|-5
Step-by-step explanation:
just rewrite the equation as k=|14+x|-5
Answer:
C
Step-by-step explanation:
