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Sindrei [870]
3 years ago
11

) 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

nsform of y(t), i.e., Y=L{y(t)}, find the equation you get by taking the Laplace transform of the differential equation (s^2-6s+9)Y-4s+22 equation editorEquation Editor =0 Solve for Y(s)= (4s-22)/(s^2-6s+9) equation editorEquation Editor write the above answer in its partial fraction decomposition, Y(s)=As+a+B(s+a)2 Y(s)= equation editorEquation Editor + equation editorEquation Editor Now, by inverting the transform, find y(t)= equation editorEquation Editor .
Mathematics
1 answer:
artcher [175]3 years ago
5 0

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)}

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Adding/subtracting polynomials <br>​
Nuetrik [128]

Answer:

g(x) - h(x) =  { - x}^{4}  - 11x

Step-by-step explanation:

g(x) =  {6x}^{2}  - 4x +  {3x}^{4}

h(x) =  {6x}^{2}  + 4 {x}^{4}  + 7x

g(x) - h(x)

({6x}^{2}  - 4x +  {3x}^{4})  - ( {6x}^{2}  + 4 {x}^{4}  + 7x)

{6x}^{2}  - 4x +  {3x}^{4}   - {6x}^{2}   -  4 {x}^{4}   -  7x

( {3x}^{4}  -  {4x}^{4}) +  ( {6x}^{2}  - 6x {}^{2} ) + ( - 4x - 7x)

{ - x}^{4}  - 11x

6 0
2 years ago
Please can I have an explanation also, I am terrible at these kinds of questions!
wlad13 [49]

Answer:

<em>The fraction of the beads that are red is</em>

Step-by-step explanation:

<u>Algebraic Expressions</u>

A bag contains red (r), yellow (y), and blue (b) beads. We are given the following ratios:

r:y = 2:3

y:b = 5:4

We are required to find r:s, where s is the total of beads in the bag, or

s = r + y + b

Thus, we need to calculate:

\displaystyle \frac{r}{r+y+b}       \qquad\qquad    [1]

Knowing that:

\displaystyle \frac{r}{y}=\frac{2}{3}      \qquad\qquad    [2]

\displaystyle \frac{y}{b}=\frac{5}{4}

Multiplying the equations above:

\displaystyle \frac{r}{y}\frac{y}{b}=\frac{2}{3}\frac{5}{4}

Simplifying:

\displaystyle \frac{r}{b}=\frac{5}{6}       \qquad\qquad    [3]

Dividing [1] by r:

\displaystyle \frac{r}{r+y+b}=\displaystyle \frac{1}{1+y/r+b/r}

Substituting from [2] and [3]:

\displaystyle \frac{r}{r+y+b}=\displaystyle \frac{1}{1+3/2+6/5}

Operating:

\displaystyle \frac{r}{r+y+b}=\displaystyle \frac{1}{\frac{10+3*5+6*2}{10}}

\displaystyle \frac{r}{r+y+b}=\displaystyle \frac{10}{10+15+12}

\displaystyle \frac{r}{r+y+b}=\displaystyle \frac{10}{37}

The fraction of the beads that are red is \mathbf{\frac{10}{37}}

8 0
3 years ago
The Math Club bought 16 graphing calculators for $765. The TI-70 model costs $40 and
Afina-wow [57]

Answer:

They bought 5

Step-by-step explanation:

65x+40y=765

65(5)+40(11)=765

5 0
2 years ago
*****50 POINTSSSS*****
defon

Answer:

<h3>Given</h3>
  • m∠REG = 78°
  • mAR = 46°
  • ER ≅ GA
<h3>Solution</h3>
  • m∠GAR = 180° - m∠REG = 180° - 78° = 102° (supplementary angles sum to 180°)
  • m∠TAR = 1/2mAR = 1/2(46°) = 23°   (tangent chord angle is half the size of intercepted arc)
  • m∠GAN = 180° - (m∠TAR + m∠GAR) = 180° - (23° + 102°) = 55° (straight angle is 180°)
  • mAG = 2m∠GAN = 2(55°) = 110°
  • mRE = mAG = 110° (as ER ≅ GA)
  • mGE = 360° - (mAG + mAR + mRE) = 360° - (110° + 46° + 110°) = 94° (full circle is 360°)
6 0
2 years ago
What is the value of x? show work
Alona [7]
2x+38=180
2x=142 (or x+x=142 from which 2x=142)
x=71 

The unknown angles are each 71

You know this answer is logical as, since two sides are 21 it hints that we are looking at an equilateral triangle, which means two of its angles will be the same
7 0
3 years ago
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