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)

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Answer:
square feet
Step-by-step explanation:
Area has units of squared length.
The equation can be (3450-1000)<span>÷7=350.</span>
<span>There are 4 long rows and 5 short rows in the theater. x represents the number of chairs in each long row and y represents the number of chairs in each short row.
So, total number of chairs in 4 long rows= 4x
Total number of chairs in 5 short rows = 5y
Total number of chairs in the theater on a normal day = 4x + 5y
When 2 chairs are added to each long row, the number of chairs will change to (x+2).
So, total number of chairs in 4 long rows will be = 4(x+2)
When 3 chairs are added to each short row, the number of chairs will change to (y+3)
So, total number of chairs in 5 short rows will be = 5(y+3)
Thus, total number of chairs in the theater in rush day = 4(x+2) + 5(y+3)
= 4x + 8 + 5y + 15
= 4x + 5y + 23
Thus we can say the number of chairs increase by 23 as compared to a normal day.
</span>
Answer:
x⁵ + 5x³ + 7x² + 35
Step-by-step explanation:
(x² + 5) (x³ + 7)
To distribute binomials (expressions with two terms), you can use the FOIL method (First, Outer, Inner, Last):
x⁵ + 7x² + 5x³ + 35
To put into standard form, rearrange from highest power to lowest power:
x⁵ + 5x³ + 7x² + 35