The sum of two numbers is five. the sum of the squares of the two numbers is thirteen. find the two numbers.

) Use the Laplace transform to solve the following initial value problem: y′′−6y′+9y=0y(0)=4,y′(0)=2 Using Y for the Laplace tra

artcher [175]

Answer:

$y(t)=2e^{3t}(2-5t)$

Step-by-step explanation:

Let Y(s) be the Laplace transform Y=L{y(t)} of y(t)

Applying the Laplace transform to both sides of the differential equation and using the linearity of the transform, we get

L{y'' - 6y' + 9y} = L{0} = 0

(*) L{y''} - 6L{y'} + 9L{y} = 0 ; y(0)=4, y′(0)=2

Using the theorem of the Laplace transform for derivatives, we know that:

$\large\bf L\left\{y''\right\}=s^2Y(s)-sy(0)-y'(0)\\\\L\left\{y'\right\}=sY(s)-y(0)$

Replacing the initial values y(0)=4, y′(0)=2 we obtain

$\large\bf L\left\{y''\right\}=s^2Y(s)-4s-2\\\\L\left\{y'\right\}=sY(s)-4$

and our differential equation (*) gets transformed in the algebraic equation

$\large\bf s^2Y(s)-4s-2-6(sY(s)-4)+9Y(s)=0$

Solving for Y(s) we get

$\large\bf s^2Y(s)-4s-2-6(sY(s)-4)+9Y(s)=0\Rightarrow (s^2-6s+9)Y(s)-4s+22=0\Rightarrow\\\\\Rightarrow Y(s)=\frac{4s-22}{s^2-6s+9}$

Now, we brake down the rational expression of Y(s) into partial fractions

$\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4s-22}{(s-3)^2}=\frac{A}{s-3}+\frac{B}{(s-3)^2}$

The numerator of the addition at the right must be equal to 4s-22, so

A(s - 3) + B = 4s - 22

As - 3A + B = 4s - 22

we deduct from here

A = 4 and -3A + B = -22, so

A = 4 and B = -22 + 12 = -10

It means that

$\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4}{s-3}-\frac{10}{(s-3)^2}$

and

$\large\bf Y(s)=\frac{4}{s-3}-\frac{10}{(s-3)^2}$

By taking the inverse Laplace transform on both sides and using the linearity of the inverse:

$\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}$

we know that

$\large\bf L^{-1}\left\{\frac{1}{s-3}\right\}=e^{3t}$

and for the first translation property of the inverse Laplace transform

$\large\bf L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=e^{3t}L^{-1}\left\{\frac{1}{s^2}\right\}=e^{3t}t=te^{3t}$

and the solution of our differential equation is

$\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=\\\\4e^{3t}-10te^{3t}=2e^{3t}(2-5t)\\\\\boxed{y(t)=2e^{3t}(2-5t)}$

5 0

3 years ago

How many $25 shirts can I buy with $175?

Nookie1986 [14]

Answer:

You can by 7 shirts if you wanted

6 0

2 years ago

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Poisson Distribution LEARNING OBJECTIVE: Calculate the probability of a poisson distribution. 3.00 The average number of bridge

ira [324]

Answer:

the required probability is 0.09

Option a) 0.09 is the correct Answer.

Step-by-step explanation:

Given that;

mean μ = 7

x = 4

the probability of exactly 4 bridge construction projects taking place at one time in this state = ?

Using the Poisson probability formula;

P( X=x ) = ( e^-μ × u^x) / x!

we substitute

P(X = 4) = (e⁻⁷ × 7⁴) / 4!

= 2.1894 / 24

= 0.0875 ≈ 0.09

Therefore the required probability is 0.09

Option a) 0.09 is the correct Answer.

5 0

2 years ago

For what value of a does<br> (9) ***<br> = 3438-1<br> -1<br> 0<br> 1<br> no solution

mario62 [17]

Answer:

solution,

$(\frac{1}{7} ) ^{3a + 3} = {343}^{a - 1} \\ or \: ( {7}^{ - 1} ) ^{3a + 3} = {(7}^{3} ) ^{a - 1} \\ or \: (7) ^{ - 3a - 3} = {(7)}^{3a - 3} \\ or \: - 3a - 3 = 3a - 33 \\ or \: - 3a - 3a = - 3 + 3 \\ or \: - 6a = 0 \\ \: a = 0$

hope this helps..

Good luck on your assignment..

3 0

3 years ago

True or False? The rule of 72 (or 69) depends on the amount of principal you are saving. *

grin007 [14]

I think answer is trueeee

3 0

3 years ago

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