A: 2x - 3y = 6

(10 points) Consider the initial value problem y′+3y=9t,y(0)=7. Take the Laplace transform of both sides of the given differenti

Rashid [163]

Answer:

The solution

$Y (s) = 9( -1 +3 t + e^{-3 t} ) + 7 e ^{-3 t}$

Step-by-step explanation:

Explanation:-

Consider the initial value problem y′+3 y=9 t,y(0)=7

Step(i):-

Given differential problem

y′+3 y=9 t

Take the Laplace transform of both sides of the differential equation

L( y′+3 y) = L(9 t)

Using Formula Transform of derivatives

L(y¹(t)) = s y⁻(s)-y(0)

By using Laplace transform formula

$L(t) = \frac{1}{S^{2} }$

Step(ii):-

Given

L( y′(t)) + 3 L (y(t)) = 9 L( t)

$s y^{-} (s) - y(0) + 3y^{-}(s) = \frac{9}{s^{2} }$

$s y^{-} (s) - 7 + 3y^{-}(s) = \frac{9}{s^{2} }$

Taking common y⁻(s) and simplification, we get

$( s + 3)y^{-}(s) = \frac{9}{s^{2} }+7$

$y^{-}(s) = \frac{9}{s^{2} (s+3}+\frac{7}{s+3}$

Step(iii):-

By using partial fractions , we get

$\frac{9}{s^{2} (s+3} = \frac{A}{s} + \frac{B}{s^{2} } + \frac{C}{s+3}$

$\frac{9}{s^{2} (s+3} = \frac{As(s+3)+B(s+3)+Cs^{2} }{s^{2} (s+3)}$

On simplification we get

9 = A s(s+3) +B(s+3) +C(s²) ...(i)

Put s =0 in equation(i)

9 = B(0+3)

B = 9/3 = 3

Put s = -3 in equation(i)

9 = C(-3)²

C = 1

Given Equation 9 = A s(s+3) +B(s+3) +C(s²) ...(i)

Comparing 'S²' coefficient on both sides, we get

9 = A s²+3 A s +B(s)+3 B +C(s²)

0 = A + C

put C=1 , becomes A = -1

$\frac{9}{s^{2} (s+3} = \frac{-1}{s} + \frac{3}{s^{2} } + \frac{1}{s+3}$

Step(iv):-

$y^{-}(s) = \frac{9}{s^{2} (s+3}+\frac{7}{s+3}$

$y^{-}(s) =9( \frac{-1}{s} + \frac{3}{s^{2} } + \frac{1}{s+3}) + \frac{7}{s+3}$

Applying inverse Laplace transform on both sides

$L^{-1} (y^{-}(s) ) =L^{-1} (9( \frac{-1}{s}) + L^{-1} (\frac{3}{s^{2} }) + L^{-1} (\frac{1}{s+3}) )+ L^{-1} (\frac{7}{s+3})$

By using inverse Laplace transform

$L^{-1} (\frac{1}{s} ) =1$

$L^{-1} (\frac{1}{s^{2} } ) = \frac{t}{1!}$

$L^{-1} (\frac{1}{s+a} ) =e^{-at}$

Final answer:-

Now the solution , we get

$Y (s) = 9( -1 +3 t + e^{-3 t} ) + 7 e ^{-3t}$

5 0

2 years ago

Find a Cartesian equation for the curve and identify it. r=8tanxsecx

ANTONII [103]

R = 8 tan(x) sec(x)

In order to solve this, we are going to use the following:
r = sqrt(x^2 + y^2) //but in this case, we don't need this.
tan(x) = Y / X
X = r cos(x)
Y = r sin(x)

(r cos x)^2 = 8(r sin x)

x^2 = 8y

3 0

2 years ago

What is the area of the figure

Semenov [28]

Answer:

Area of the figure = 36 in²

Step-by-step explanation:

Area of the given composite figure = Area of the rectangle ABCE - Area of triangle CDE

Area of rectangle ABCE = Length × Width

= AB × AC

= 6 × 8

= 48 in²

Area of triangle CDE = $\frac{1}{2}(\text{Base})\times (\text{Height})$

= $\frac{1}{2}(6)(4)$

= 12 in²

Now area of the composite figure = 48 - 12

= 36 square inch

4 0

2 years ago

Please help need it quick will give brainliest thanks

Norma-Jean [14]

Use the Pythagorean theorem:

$x^2+8^2=11^2\\\\x^2+64=121\ \ \ |-64\\\\x^2=57\to x=\sqrt{57}\\\\x\approx7.5$

Answer: C. 7.5

8 0

2 years ago

Read 2 more answers

Just help me plssssssssss

Komok [63]

Answer:

The perimeter of triangle is 30ft.

Step-by-step explanation:

First you have to find the value of a using Pythogoras Theorem, a² + b² = c² where a and b is the sides and c is the hypotenuse :

${a}^{2} + {b}^{2} = {c}^{2}$

${a}^{2} = {c}^{2} - {b}^{2}$

$a = \sqrt{ {c}^{2} - {b}^{2} }$

Let b = 12,

Let c = 13,

$a = \sqrt{ {13}^{2} - 12 {?}^{2} }$

$a = \sqrt{25}$

$a = 5feet$

Next you have to find the perimeter of triangle by adding all the sides together :

$perimeter = 5 + 12 + 13$

$perimeter = 30feet$

7 0

2 years ago