The sum of three consecutive numbers is 72 .what are the smallest of these numbers?

Mathematics

2 answers:

Vikentia [17]3 years ago

4 0

The smallest of these consecutive numbers would be 23 because 23+24+25 = 72.

Step2247 [10]3 years ago

3 0

The answer is 23, 24, 25

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Use the laplace transform to solve the given initial-value problem. y'' − 4y' + 4y = t3e2t, y(0) = 0, y'(0) = 0

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The Laplace of the given equation is $y=-\frac{15e^2}{8}e^{2t}+\left(-3e^2+\frac{15e^2}{4}\right)e^{2t}t+\frac{e^2t^4}{4}+e^2t^3+\frac{9e^2t^2}{4}+3e^2t+\frac{15e^2}{8}$ .

According to the statement

we have given that the equation and we have to solve this problem with the help of the Laplace transform.

So, According to the statement

the given equation is

y'' − 4y' + 4y = t^3e^2t, y(0) = 0, y'(0) = 0

And the Laplace transform is an integral transform method which is particularly useful in solving linear ordinary differential equations.

Firstly fill the value of the Laplace of the second order derivative and put x = 0 in the equation

And same it with the first order of the derivative.

And then the Laplace of the equation become

$y=-\frac{15e^2}{8}e^{2t}+\left(-3e^2+\frac{15e^2}{4}\right)e^{2t}t+\frac{e^2t^4}{4}+e^2t^3+\frac{9e^2t^2}{4}+3e^2t+\frac{15e^2}{8}$

So, The Laplace of the given equation is $y=-\frac{15e^2}{8}e^{2t}+\left(-3e^2+\frac{15e^2}{4}\right)e^{2t}t+\frac{e^2t^4}{4}+e^2t^3+\frac{9e^2t^2}{4}+3e^2t+\frac{15e^2}{8}$

Learn more about Laplace transform here

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Step-by-step explanation: