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2 years ago

The partial fraction decomposition of 9 x 11 6 x 2 23 x 21 can be written in the form of f ( x ) 2 x 3 g ( x ) 3 x 7 , where

Mathematics

1 answer:

svlad2 [7]2 years ago

6 0

Answer:

$f(x) = -1$

$g(x) =6$

$\frac{-1}{2x + 3} + \frac{6}{3x + 7}$

Step-by-step explanation:

The question is unreadable, however the real polynomial is:

The polynomial fraction is:

$P = \frac{9x + 11}{6x^2 + 23x + 21}$

And the decomposition is:

$\frac{f(x)}{2x + 3} + \frac{g(x)}{3x + 7}$

The solution is as follows:

$P = \frac{f(x)}{2x + 3} + \frac{g(x)}{3x + 7}$

Substitute the expression for P

$\frac{9x + 11}{6x^2 + 23x + 21} = \frac{f(x)}{2x + 3} + \frac{g(x)}{3x + 7}$

Expand the numerator of the polynomial

$\frac{9x + 11}{(2x + 3)(3x + 7)} = \frac{f(x)}{2x + 3} + \frac{g(x)}{3x + 7}$

Take LCM

$\frac{9x + 11}{(2x + 3)(3x + 7)} = \frac{f(x)*(3x + 7) + g(x)*(2x + 3)}{(2x + 3)(x + 7)}$

Cancel out both denominators

$9x + 11} = f(x)*(3x + 7) + g(x)*(2x + 3)$

Represent f(x) as A and g(x) as B

$9x + 11} = A*(3x + 7) + B*(2x + 3)$

Open bracket

$9x + 11} = 3Ax + 7A + 2Bx + 3B$

$9x + 11} = 3Ax + 2Bx+ 7A + 3B$

$9x + 11} = (3A + 2B)x+ 7A + 3B$

By comparison:

$3A + 2B = 9$ ---- (1)

$7A + 3B =11$ ---- (2)

Make B the subject in (1)

$B = \frac{9 - 3A}{2}$

Substitute $B = \frac{9 - 3A}{2}$ in (2)

$7A + 3(\frac{9 - 3A}{2}) = 11$

Multiply through by 2

$2*7A +2* 3(\frac{9 - 3A}{2}) = 11*2$

$14A + 3(9 - 3A) = 22$

$14A + 27 - 9A = 22$

Collect Like Terms

$14A - 9A = 22-27$

$5A = -5$

$A = \frac{-5}{5}$

$A = -1$

Recall that:

$B = \frac{9 - 3A}{2}$

$B = \frac{9 - 3 * -1}{2}$

$B = \frac{9 + 3}{2}$

$B = \frac{12}{2}$

$B = 6$

A = -1 and B = 6

$3A + 2B = 9$ ---- (1)

$7A + 3B =11$ ---- (2)

So:

$f(x) = A$

$f(x) = -1$

$g(x) =B$

$g(x) =6$

And the decomposition is:

$\frac{-1}{2x + 3} + \frac{6}{3x + 7}$

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3 years ago

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3 years ago

) 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

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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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